Mathematics-Part-2---Class-12

THREE DIMENSIONAL GEOMETRY 475 Hence, the required shortest distance is d = PQ = ST | cos θ | or d = Cartesian form The shortest distance between the lines l1 : x − x1 = y − y1 = z − z1 a1 b1 c1 and l2 : x − x2 = y − y2 = z − z2 a2 b2 c2 x2 − x1 y2 − y1 z2 − z1 a1 b1 c1 a2 b2 c2 is (b1c2 − b2c1 )2 + (c1a2 − c2a1 )2 + (a1b2 − a2b1 )2 11.5.2 Distance between parallel lines If two lines l1 and l2 are parallel, then they are coplanar. Let the lines be given by ... (1) and … (2) wa2heisret,hea1piossitthieonpovseicttioornovfeactpoorinotf a point S on l1 and T on l Fig 11.9. 2 As l1, l2 are coplanar, if the foot of the perpendicular from T on the line l is P, then the distance between the 1 lines l1 and l2 = | TP |.   Let θ be the angle between the vectors ST and b . Fig 11.9 Then ... (3)   b × ST = where nˆ is the unit vector perpendicular to the plane of the lines l1 and l2. But  a2 − a1 ST = 2019-20

476 MATHEMATICS Therefore, from (3), we get  b × (a2 − a1) =  (since PT = ST sin θ) | b | PT nˆ (as | nˆ | = 1) i.e., |  × (a2 − a1)| =  b | b | PT ⋅1 Hence, the distance between the given parallel lines is d= Example 11 Find the shortest distance between the lines l1 and l2 whose vector equations are r = iˆ + ˆj + λ (2 iˆ − ˆj + kˆ ) ... (1) and r = 2 iˆ + ˆj − kˆ + µ (3 iˆ −5 ˆj + 2 kˆ ) ... (2) Solution Comparing (1) and (2) with r = a1 +  and r =  + µ  respectively, we get a2 = λb1 a2 b2 b1 a1 Therefore a2 = iˆ + ˆj , b1 = 2 iˆ − ˆj + kˆ and − a1 = 2 iˆ + ˆj – kˆ and  × b2 = b2 = 3 iˆ – 5 ˆj + 2 kˆ iˆ − kˆ ( 2 iˆ − ˆj + kˆ ) × ( 3 iˆ − 5 ˆj + 2 kˆ ) iˆ ˆj kˆ = 2 −1 1 = 3 iˆ − ˆj − 7 kˆ 3 −5 2  So |b1 × b2 | = 9 +1 + 49 = 59 Hence, the shortest distance between the given lines is given by d=   .  −  ) = | 3− 0 + 7 | = 10 ( b1 × b2) (a2 a1 | b1 × b2 | 59 59 Example 12 Find the distance between the lines l1 and l2 given by r = iˆ + 2 ˆj − 4 kˆ + λ ( 2 iˆ + 3 ˆj + 6 kˆ ) and r = 3iˆ + 3 ˆj − 5 kˆ + µ ( 2 iˆ + 3 ˆj + 6 kˆ ) 2019-20

THREE DIMENSIONAL GEOMETRY 477 Solution The two lines are parallel (Why? ) We have  a2 b a1 = iˆ + 2 ˆj − 4 kˆ , = 3iˆ + 3 ˆj − 5 kˆ and = 2iˆ + 3 ˆj + 6 kˆ Therefore, the distance between the lines is given by d=  × (a2 − a1 ) = iˆ ˆj kˆ b 23 6 2 1 −1 |b | 4 + 9 + 36 or = | − 9iˆ + 14 ˆj − 4 kˆ | = 293 = 293 49 49 7 EXERCISE 11.2 1. Show that the three lines with direction cosines 12 , −3 , −4 ; 4 , 12 , 3 ; 3 , −4 , 12 are mutually perpendicular. 13 13 13 13 13 13 13 13 13 2. Show that the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6). 3. Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (– 1, – 2, 1), (1, 2, 5). 4. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3 iˆ + 2 ˆj − 2 kˆ . 5. Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2 iˆ − j + 4 kˆ and is in the direction iˆ + 2 ˆj − kˆ . 6. Find the cartesian equation of the line which passes through the point (– 2, 4, – 5) and parallel to the line given by x + 3 = y − 4 = z + 8 . 356 7. The cartesian equation of a line is x − 5 = y + 4 = z − 6 . Write its vector form. 372 8. Find the vector and the cartesian equations of the lines that passes through the origin and (5, – 2, 3). 2019-20

478 MATHEMATICS 9. Find the vector and the cartesian equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6). 10. Find the angle between the following pairs of lines: (i) r = 2iˆ − 5 ˆj + kˆ + λ (3iˆ + 2 ˆj + 6 kˆ) and r = 7iˆ − 6 kˆ + µ(iˆ + 2 ˆj + 2 kˆ) (ii) r = 3iˆ + ˆj − 2 kˆ + λ (iˆ − ˆj − 2kˆ) and r = 2iˆ − ˆj − 56 kˆ + µ (3 iˆ − 5 ˆj − 4kˆ) 11. Find the angle between the following pair of lines: (i) x − 2 = y −1 = z + 3 and x + 2 = y − 4 = z − 5 2 5 −3 −1 8 4 (ii) x = y = z and x − 5 = y − 2 = z − 3 221 4 1 8 12. Find the values of p so that the lines 1 − x = 7 y − 14 = z − 3 3 2p 2 and 7 − 7x = y − 5 = 6 − z are at right angles. 3p 1 5 13. Show that the lines x − 5 = y + 2 = z and x = y = z are perpendicular to 7 −5 1 123 each other. 14. Find the shortest distance between the lines r = (iˆ + 2 ˆj + kˆ) + λ (iˆ − ˆj + kˆ) and r = 2 iˆ − ˆj − kˆ + µ (2 iˆ + ˆj + 2 kˆ) 15. Find the shortest distance between the lines x +1= y +1= z +1 and x−3= y−5= z−7 7 −6 1 1 −2 1 16. Find the shortest distance between the lines whose vector equations are r = (iˆ + 2 ˆj + 3 kˆ) + λ (iˆ − 3 ˆj + 2 kˆ) and r = 4 iˆ + 5 ˆj + 6 kˆ + µ (2 iˆ + 3 ˆj + kˆ) 17. Find the shortest distance between the lines whose vector equations are r = (1 − t) iˆ + (t − 2) ˆj + (3− 2 t) kˆ and r = (s +1) iˆ + (2s − 1) ˆj − (2s + 1) kˆ 2019-20

THREE DIMENSIONAL GEOMETRY 479 11.6 Plane A plane is determined uniquely if any one of the following is known: (i) the normal to the plane and its distance from the origin is given, i.e., equation of a plane in normal form. (ii) it passes through a point and is perpendicular to a given direction. (iii) it passes through three given non collinear points. Now we shall find vector and Cartesian equations of the planes. 11.6.1 Equation of a plane in normal form Considera plane whose perpendicular distance from the origin is d (d ≠ 0). Fig 11.10. If ON is the norm al from the origin to the plane, and nˆ is the unit normal vector . Then ON = d nˆ . Let P beany along ON Z point on the plane. Therefore, NP is perpendicularto ON . Therefore, NP ⋅ ON = 0 ... (1) thenLNetPr=berth−e P(x,y,z) position vectorofthepoint P, r d nˆ (as ON + NP = OP ) dN Therefore, (1) becomes OY (r − d n∧ ) ⋅ d ∧ = 0 X n or (r − d n∧) ⋅ n∧ = 0 (d ≠ 0) Fig 11.10 or r ⋅ n∧ − d n∧ ⋅ n∧ = 0 i.e., r ⋅ n∧ = d (as ∧ ⋅ ∧ = 1) … (2) n n This is the vector form of the equation of the plane. Cartesian form Equation (2) gives the vector equation of a plane, where nˆ is the unit vector normal to the plane. Let P(x, y, z) be any point on the plane. Then  r = OP = x iˆ + y ˆj + z kˆ Let l, m, n be the direction cosines of nˆ . Then nˆ = l iˆ + m ˆj + n kˆ 2019-20

480 MATHEMATICS Therefore, (2) gives (x iˆ + y ˆj + z kˆ) ⋅ (l iˆ + m ˆj + n kˆ) = d i.e., lx + my + nz = d ... (3) This is the cartesian equation of the plane in the normal form. Note Equation (3) shows that if r ⋅ (a iˆ + b ˆj + c kˆ) = d is the vector equation of a plane, then ax + by + cz = d is the Cartesian equation of the plane, where a, b and c are the direction ratios of the normal to the plane. 6 Example 13 Find the vector equation of the plane which is at a distance of 29 from the origin and its normal vector from the origin is 2iˆ − 3 ˆj + 4kˆ . Also find its cartesian form.  Solution Let n = 2 iˆ −3 ˆj + 4 kˆ . Then nˆ = n = 2 iˆ − 3 ˆj + 4 kˆ = 2 iˆ − 3 ˆj + 4 kˆ | n | 4 + 9 + 16 29 Hence, the required equation of the plane is r ⋅ 2 iˆ + −3 ˆj + 4 kˆ  = 6 29 29 29  29 Example 14 Find the direction cosines of the unit vector perpendicular to the plane r ⋅ (6 iˆ − 3 ˆj − 2 kˆ) + 1 = 0 passing through the origin. Solution The given equation can be written as ... (1) r ⋅ ( − 6 iˆ + 3 ˆj + 2 kˆ ) = 1 Now | − 6 iˆ + 3 ˆj + 2 kˆ | = 36 + 9 + 4 = 7 Therefore, dividing both sides of (1) by 7, we get r ⋅  − 6 iˆ + 3 ˆj + 2 kˆ  = 1  7 7 7  7 form r ⋅ nˆ which is the equation of the plane in the = d . This shows that nˆ = − 6 iˆ + 3 ˆj + 2 kˆ is a unit vector perpendicular to the 77 7 plane through the origin. Hence, the direction cosines of nˆ are − 6 , 3 , 2 . 7 77 2019-20

THREE DIMENSIONAL GEOMETRY 481 Example 15 Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin. Solution Since the direction ratios of the normal to the plane are 2, –3, 4; the direction cosines of it are 2 −3 4 2 , −3 , 4 , , 22 + (−3)2 + 42 , i.e., 29 29 29 22 + (− 3)2 + 42 22 + (−3)2 + 42 Hence, dividing the equation 2x – 3y + 4z – 6 = 0 i.e., 2x – 3y + 4z = 6 throughout by 29 , we get 2 x + −3 y + 4 z = 6 29 29 29 29 This is of the form lx + my + nz = d, where d is the distance of the plane from the 6 origin. So, the distance of the plane from the origin is . 29 Example 16 Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x – 3y + 4z – 6 = 0. Solution Let the coordinates of the foot of the perpendicular P from the origin to the plane is (x1, y1, z1) (Fig 11.11). Z Then, the direction ratios of the line OP are x1, y1, z1. P(x1, y1, z1) Writing the equation of the plane in the normal form, we have 2 x− 3 y + 4 z = 6 O Y 29 29 29 29 where, 2 , −3 , 4 are the direction X 29 29 29 cosines of the OP. Fig 11.11 Since d.c.’s and direction ratios of a line are proportional, we have x1 = y1 = z1 =k 2 −3 4 29 29 29 i.e., x1 = 2k −3k , z1 = 4k 29 , y1 = 29 29 2019-20

482 MATHEMATICS 6 Substituting these in the equation of the plane, we get k = . 29 Hence, the foot of the perpendicular is  12 , −18 , 24  .  29 29 29  Note If d is the distance from the origin and l, m, n are the direction cosines of the normal to the plane through the origin, then the foot of the perpendicular is (ld, md, nd). 11.6.2 Equation of a plane perpendicular to a given vector and passing through a given point In the space, there can be many planes that are perpendicular to the given vector, but through a given point P(x1, y1, z1), only one such plane exists (see Fig 11.12). vectoLretaa plane pass through a point A withposition Fig 11.12 and perpendicular to the vector N . Let r be the position vector of any point P(x, y, z) in the plane. (Fig 11.13).  Then the point P lies in the planeif and only if AP is perpendicular to N .a)i.⋅eN., AP . N = 0. But AP = r − a . Therefore, (r − =0 … (1) This is the vector equation of the plane. Cartesian form Let the given point Abe (x1, y1, z1), P be (x, y, z) Fig 11.13 and direction ratios of N are A, B and C. Then, a = x1 iˆ + y1 ˆj + z1 kˆ, r = xiˆ + y ˆj + z kˆ and  = A iˆ + B ˆj + C kˆ N Now (r – a) ⋅  = 0 N So ( x − x1 )iˆ+ ( y − y1 ) ˆj +( z − z1 ) kˆ⋅(Aiˆ + B ˆj + Ckˆ)=0 i.e. A (x – x1) + B (y – y1) + C (z – z1) = 0 Example 17 Find the vector and cartesian equations of the plane which passes through the point (5, 2, – 4) and perpendicular to the line with direction ratios 2, 3, – 1. 2019-20

THREE DIMENSIONAL GEOMETRY 483 Solution We have the position vector of point (5, 2, – 4) as a =5iˆ + 2 ˆj − 4kˆ and the normal vector  perpendicular to the plane as  = 2iˆ +3 ˆj − kˆ N N Therefore, the vector equation of the plane is given by (r − a)  = 0 .N or [r − (5 iˆ + 2 ˆj − 4 kˆ)] ⋅ (2 iˆ + 3 ˆj − kˆ) = 0 ... (1) Transforming (1) into Cartesian form, we have [(x – 5)iˆ + ( y − 2) ˆj + (z + 4) kˆ] ⋅ (2 iˆ + 3 ˆj − kˆ) = 0 or 2(x −5) +3( y − 2) −1(z + 4) = 0 i.e. 2x + 3y – z = 20 which is the cartesian equation of the plane. 11.6.3 Equation of a plane passing through three non collinear points Lceret sRp,eSctainvdelTy be three non collinear points on the plane with position vectors a  and (Fig 11.14). ,b Z (RS X RT) R P ra S c T Ob Y X   Fig 11.14   is The vectors RS and RT are in the given Rp,laSnaen. dTTh.eLreeftorre,btehtehveepcotosirtioRnSv×ecRtoTr perpendicular to the plane containing points of any point P in the plane. Therefore, the equation of the plane passing through R and perpendicular to the vector RS × RT is (r − a) ⋅   = 0 (RS× RT) or (r – a).[(b – a)×(c – a)] = 0 … (1) 2019-20

484 MATHEMATICS This is the equation of the plane in vector form passing through three noncollinear points. Note Why was it necessary to say that the three points R S had to be non collinear? If the three points were on the same line, then there will be many planes that will contain them (Fig 11.15). These planes will resemble the pages of a book where the T line containing the points R, S and T are members in the binding of the book. Cartesian form Fig 11.15 Let (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be the coordinates of the points R, S and T rveescpteocrtirve. lTy.heLnet (x, y, z) be the coordinates of any point P on the plane with position  RP = (x – x1) iˆ + (y – y1) ˆj + (z – z1) kˆ RS = (x2 – x1) iˆ + (y2 – y1) ˆj + (z2 – z1) kˆ  = (x3 – x1) iˆ + (y3 – y1) ˆj + (z3 – z1) kˆ RT Substituting these values in equation (1) of the vector form and expressing it in the form of a determinant, we have x − x1 y − y1 z − z1 x2 − x1 y2 − y1 z2 − z1 = 0 x3 − x1 y3 − y1 z3 − z1 which is the equation of the plane in Cartesian form passing through three non collinear points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3). Example 18 Find the vector equations of the plane passing through the points R (2, 5, – 3), S (– 2, – 3, 5) and T(5, 3,– 3).  Solution Let a = 2iˆ + 5 ˆj − 3kˆ , b = − 2 iˆ − 3 ˆj + 5 kˆ , c = 5iˆ + 3 ˆj − 3kˆ  b Then the vector equation of the plane passing through a , and c and is given by (r − a)   0 ] (Why?) or (r − a) ⋅⋅[((RbS−×aR)×T()c=− a) =0 i.e. [r − (2iˆ + 5 ˆj − 3kˆ)]⋅[(−4iˆ −8 ˆj +8 kˆ) × (3iˆ − 2 ˆj)] = 0 2019-20

THREE DIMENSIONAL GEOMETRY 485 11.6.4 Intercept form of the equation of a plane In this section, we shall deduce the equation of a plane in terms of the intercepts made by the plane on the coordinate axes. Let the equation of the plane be Ax + By + Cz + D = 0 (D ≠ 0) ... (1) Let the plane make intercepts a, b, c on x, y and z axes, respectively (Fig 11.16). Hence, the plane meets x, y and z-axes at (a, 0, 0), (0, b, 0), (0, 0, c), respectively. Therefore −D Aa + D = 0 or A = a −D Bb + D = 0 or B = b −D Fig 11.16 Cc + D = 0 or C = c Substituting these values in the equation (1) of the plane and simplifying, we get x+y+z =1 ... (1) abc which is the required equation of the plane in the intercept form. Example 19 Find the equation of the plane with intercepts 2, 3 and 4 on the x, y and z-axis respectively. Solution Let the equation of the plane be x+y+z =1 ... (1) abc Here a = 2, b = 3, c = 4. Substituting the values of a, b and c in (1), we get the required equation of the plane as x + y + z = 1 or 6x + 4y + 3z = 12. 234 11.6.5 Plane passing through the intersection of two given planes Lre⋅ tnˆ1π=1 da1nadndπr2 be two planes with equations ⋅ nˆ2 = d respectively. The position 2 vector of any point on the line of intersection must satisfy both the equations (Fig 11.17). Fig 11.17 2019-20

486 MATHEMATICS  If t is the position vector of a point on the line, then  t ⋅ nˆ1 = d and t ⋅ nˆ2 =d 1 2 Therefore, forall real values of λ, we have  t ⋅ (nˆ1 + λnˆ2 ) = d1 + λd2 SthiantcHiefeatnncyiesv,aetrhcbteiotrreaqrryu,asitatiotsinasftiirsefs⋅ie(bnso1ft+horλthanen2y)e=qpudoa1itn+itoλondns2πtrh1eeapnlriednseπe.n2,tsit a plane π3 which is such also satisfies the equation π i.e., any plane passing through the i=ntedr1saencdtiorn⋅ no2f the planes 3 r ⋅ (n1 + rλ⋅nn21) = d2 has the equation = d + λd ... (1) 1 2 Cartesian form In Cartesian system, let n1 = A1 iˆ + B2 ˆj + C1 kˆ and n2 = A2 iˆ + B2 ˆj + C2 kˆ r = x iˆ + y ˆj + z kˆ Then (1) becomes x (A1 + λA2) + y (B1 + λB2) + z (C1 + λC2) = d1 + λd2 or (A1 x + B1y + C1z – d1) + λ (A2 x + B2 y + C2 z – d2) = 0 ... (2) which is the required Cartesian form of the equation of the plane passing through the intersection of the given planes for each value of λ. Example 20 Find the vector equation of the plane passing through the intersection of the planes r ⋅ (iˆ + ˆj + kˆ) = 6 and r ⋅ (2iˆ + 3 ˆj + 4kˆ) = − 5, and the point (1, 1, 1). Solution Here, n1 = iˆ + ˆj + kˆ and n2 = 2iˆ + 3 ˆj + 4kˆ; and d1 = 6 and d2 = –5 Hence, using the relation r ⋅ (n1 + λn2 ) = d1 + λd2 , we get r ⋅[iˆ + ˆj + kˆ + λ (2iˆ + 3 ˆj + 4 kˆ )] = 6 − 5λ or r ⋅[(1 + 2λ) iˆ + (1+ 3λ) ˆj + (1+ 4λ)kˆ] = 6 − 5λ … (1) where, λ is some real number. 2019-20

THREE DIMENSIONAL GEOMETRY 487 Taking r = x iˆ + y ˆj + z kˆ , we get (x iˆ + y ˆj + z kˆ)⋅[(1+ 2λ) iˆ + (1+ 3λ) ˆj + (1+ 4λ)kˆ]= 6 −5λ or (1 + 2λ ) x + (1 + 3λ) y + (1 + 4λ) z = 6 – 5λ ... (2) or (x + y + z – 6 ) + λ (2x + 3y + 4 z + 5) = 0 Given that the plane passes through the point (1,1,1), it must satisfy (2), i.e. (1 + 1 + 1 – 6) + λ (2 + 3 + 4 + 5) = 0 3 or λ = 14 Putting the values of λ in (1), we get r 1+ 3  iˆ + 1+ 9  ˆj + 1+ 6  kˆ = 6 −15 7  14  7  14 or r  10 iˆ + 23 ˆj + 13 kˆ  = 69  7 14 7  14 or r ⋅ (20iˆ + 23 ˆj + 26 kˆ) = 69 which is the required vector equation of the plane. 11.7 Coplanarity of Two Lines Let the given lines be a1  and r = a2 + λb1 r = + µb2 aa12 ... (1) ... (2) The line (1) passes through the point, say A, with position vector and is parallel  and is parallel to b1 . The line (2) passes through the point, say B with position vector to b2 . Thus,  =  −    AB a2 a1 The given lines are coplanar if and only if AB is perpendicular to b1 ×b2 . i.e.      AB.(b1 × b2 ) = 0 or (a2 − a1 ) ⋅ (b1 ×b2 ) =0 Cartesian form Let (x1, y1, z1) and (x2, y2, z2) be the coordinates of the points A and B respectively. 2019-20

488 MATHEMATICS b1, anda2, b2, c2 be the direction  Let a1, c1 ratios of b1 and b2 , respectively. Then A B= ( x2 − x1)iˆ + (kˆya2n−dy1b)2 ˆj + (z2 − z1)kˆ b1 = a1 iˆ + b1 ˆj + c1 = a2iˆ+ b2ˆj +c2 kˆ ( )The given lines are coplanar if and only if AB⋅ b1 ×b2 = 0 . In the cartesian form, it can be expressed as x2 − x1 y2 − y1 z2 − z1 ... (4) a1 b1 c1 = 0 a2 b2 c2 Example 21 Show that the lines x +3 = y −1 = z −5 and x +1 = y − 2 = z −5 are coplanar. –3 1 5 –1 2 5 Solution Here, x1 = – 3, y1 = 1, z1 = 5, a1 = – 3, b1 = 1, c1 = 5 x2 = – 1, y2 = 2, z2 = 5, a2 = –1, b2 = 2, c2 = 5 Now, consider the determinant x2 − x1 y2 − y1 z2 − z1 2 1 0 a1 b1 c1 = −3 1 5 = 0 a2 b2 c2 −1 2 5 Therefore, lines are coplanar. 11.8 Angle between Two Planes Definition 2 The angle between two planes is defined as the angle between their normals (Fig 11.18 (a)). Observe that if θ is an angle between the two planes, then so is 180 – θ (Fig 11.18 (b)). We shall take the acute angle as the angles between two planes. Fig 11.18 2019-20

THREE DIMENSIONAL GEOMETRY 489 If n1 and n2 are normals to the planes and θ be the angle between the planes r ⋅  = d1 and r .  = d2 . n1 n2 Then θ is the angle between the normals to the planes drawn from some common point. We have, cos θ =  Note The planes are perpendicular to each other if  .  = 0 and parallel if  n1 n2 n1 n2 is parallel to . Cartesian form Let θ be the angle between the planes, A1 x + B1 y + C1z + D1 = 0 and A2x + B2 y + C2 z + D2 = 0 The direction ratios of the normal to the planes are A1, B1, C1 and A2, B2, C2 respectively. A1 A2 + B1 B2 + C1 C2 Therefore, cos θ = A12 + B12 + C12 A 2 + B22 + C22 2 Note 1. If the planes are at right angles, then θ = 90o and so cos θ = 0. Hence, cos θ = A1A2 + B1B2 + C1C2 = 0. 2. If the planes are parallel, then A1 = B1 = C1 . A2 B2 C2 Example 22 Find the angle between the two planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7 using vector method. Solution The angle between two planes is the angle between their normals. From the equation of the planes, the normal vectors are  = 2 iˆ + ˆj − 2 kˆ and  = 3 iˆ − 6 ˆj − 2 kˆ N1 N2 Therefore cos θ = 4 Hence =  21  θ = cos – 1  4   21  2019-20

490 MATHEMATICS Example 23 Find the angle between the two planes 3x – 6y + 2z = 7 and 2x + 2y – 2z =5. Solution Comparing the given equations of the planes with the equations We get A1 x + B1 y + C1 z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0 A1 = 3, B1 = – 6, C1 = 2 A2 = 2, B2 = 2, C2 = – 2 cos θ = 3 × 2 + (−6) (2) + (2) (−2) (32 + (− 6)2 + (−2)2 ) (22 + 22 + (−2)2 ) = −10 = 5 =5 3 7×2 3 7 3 21 Therefore, 5 3 θ = cos-1  21  11.9 Distance of a Point from a Plane Vector form Cro⋅ nnˆsi=dder(Faigp1o1in.1t9P). with position vector a and a plane π1 whose equation is ZZ p2 p1 p 2 P Q p1 N P a N’ d a O d O N’ Y YN X X (b) (a) Fig 11.19 Consider a plane π2 through Pr p−araall)e⋅lnˆto=t0he plane π1. The unit vector normal to r ⋅ nˆ = a ⋅ nˆ π is nˆ . Hence, its equation is ( 2 i.e., Thus, the distance ON′ of this plane from the origin is |a ⋅ nˆ | . Therefore, the distance PQ from the plane π is (Fig. 11.21 (a)) i.e., 1 ON – ON′ = | d – a ⋅ nˆ | 2019-20

THREE DIMENSIONAL GEOMETRY 491 which is the length of the perpendicular from a point to the given plane. We may establish the similar results for (Fig 11.19 (b)). Note r ⋅   is N N 1. If the equation of the plane π2 is in the form | = d, where is normal to the plane, then the perpendicular distance a ⋅  − d |. N |N| 2. The length of the perpendicular from origin O to the plane r ⋅  = d is | d | N |N| (since a = 0). Cartesian form  Let P(x1, y1, z1) be the given point with position vector a and Ax + By + Cz = D be the Cartesian equation of the given plane. Then a = x1 iˆ + y1 ˆj + z1 kˆ  N = A iˆ + B ˆj + C kˆ Hence, from Note 1, the perpendicular from P to the plane is ( x1 iˆ + y1 ˆj + z1 kˆ ) ⋅ ( A iˆ + B ˆj + C kˆ ) − D A2 + B2 + C2 = A x1 + B y1 + C z1 − D A2 + B2 + C2 Example 24 Find the distance of a point (2, 5, – 3) from the plane r ⋅ ( 6 iˆ − 3 ˆj + 2 kˆ ) = 4  Solution Here, a = 2 iˆ + 5 ˆj − 3 kˆ , N = 6 iˆ − 3 ˆj + 2 kˆ and d = 4. Therefore, the distance of the point (2, 5, – 3) from the given plane is | (2 iˆ +5 ˆj − 3 kˆ) ⋅ (6 iˆ − 3 ˆj + 2 kˆ) − 4| | 12 − 15 − 6 − 4| = 13 | 6 iˆ − 3 ˆj + 2 kˆ | 36 + 9 + 4 7 = 2019-20

492 MATHEMATICS 11.10 Angle between a Line and a Plane Definition 3 The angle between a line and a plane is the complement of the angle between the line and normal to the plane (Fig 11.20). r =Vaec+tλorbfoarnmd If the equation of the line is the equation of the plane is r ⋅ n = d . Then the angle θ between the line and the normal to the plane is Fig 11.20 cos θ = |  ⋅ |nn | b ⋅ b| and so the angle φ between the line and the plane is given by 90 – θ, i.e., sin (90 – θ) = cos θ i.e. sin φ =  ⋅ |nn or φ = sin–1 b ⋅n b bn | b| | Example 25 Find the angle between the line x +1 y = z−3 = 236 and the plane 10 x + 2y – 11 z = 3. Solution Let θ be the angle between the line and the normal to the plane. Converting the given equations into vector form, we have r = ( – iˆ + 3 kˆ) + λ ( 2 iˆ + 3 ˆj + 6 kˆ ) and r ⋅ ( 10 iˆ + 2 ˆj − 11 kˆ ) = 3 Here  = 2 iˆ + 3 ˆj + 6 kˆ and n = 10 iˆ + 2 ˆj − 11 kˆ b (2 iˆ + 3 ˆj + 6 kˆ) ⋅ (10 iˆ + 2 ˆj − 11 kˆ) sin φ = 22 + 32 + 62 102 + 22 + 112 = − 40 = −8 8 or φ = sin −1  8  7 × 15 21 = 21  21  2019-20

THREE DIMENSIONAL GEOMETRY 493 EXERCISE 11.3 1. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin. (a) z = 2 (b) x + y + z = 1 (c) 2x + 3y – z = 5 (d) 5y + 8 = 0 2. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 3 iˆ + 5 ˆj − 6 kˆ. 3. Find the Cartesian equation of the following planes: (a) r ⋅ (iˆ + ˆj − kˆ) = 2 (b) r ⋅ (2iˆ + 3 ˆj − 4 kˆ) = 1 (c) r ⋅[(s − 2t) iˆ + (3 − t) ˆj + (2 s + t ) kˆ] = 15 4. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. (a) 2x + 3y + 4z – 12 = 0 (b) 3y + 4z – 6 = 0 (c) x + y + z = 1 (d) 5y + 8 = 0 5. Find the vector and cartesian equations of the planes (a) that passes through the point (1, 0, – 2) and the normal to the plane is iˆ + ˆj − kˆ. (b) that passes through the point (1,4, 6) and the normal vector to the plane is iˆ − 2 ˆj + kˆ. 6. Find the equations of the planes that passes through three points. (a) (1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3) (b) (1, 1, 0), (1, 2, 1), (– 2, 2, – 1) 7. Find the intercepts cut off by the plane 2x + y – z = 5. 8. Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane. 9. Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1). 10. Find the vector equation of the plane passing through the intersection of the planes r .( 2 iˆ + 2 ˆj − 3 kˆ ) = 7 , r .( 2 iˆ + 5 ˆj + 3 kˆ ) = 9 and through the point (2, 1, 3). 11. Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0. 2019-20

494 MATHEMATICS 12. Find the angle between the planes whose vector equations are r ⋅ (2 iˆ + 2 ˆj − 3 kˆ) = 5 and r ⋅ (3 iˆ − 3 ˆj + 5 kˆ) = 3. 13. In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. (a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0 (b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0 (c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0 (d) 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0 (e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0 14. In the following cases, find the distance of each of the given points from the corresponding given plane. Point Plane (a) (0, 0, 0) 3x – 4y + 12 z = 3 (b) (3, – 2, 1) 2x – y + 2z + 3 = 0 (c) (2, 3, – 5) x + 2y – 2z = 9 (d) (– 6, 0, 0) 2x – 3y + 6z – 2 = 0 Miscellaneous Examples Example 26 A line makes angles α, β, γ and δ with the diagonals of a cube, prove that 4 cos2 α + cos2 β + cos2 γ + cos2 δ = 3 Solution A cube is a rectangular parallelopiped having equal length, breadth and height. Let OADBFEGC be the cube with each side of length a units. (Fig 11.21) The four diagonals are OE, AF, BG and CD. Z The direction cosines of the diagonal OE which C(0, 0, a) is the line joining two points O and E are F(0, a, a) (a, 0, a) G E(a,a,a) a−0 a−0 a−0 O Y ,, B(0, a, 0) a2 + a2 + a2 a2 + a2 + a2 a2 + a2 + a2 11 1 A(a, 0, 0) D(a, a, 0) i.e., 3 , 3 , 3 X Fig 11.21 2019-20

THREE DIMENSIONAL GEOMETRY 495 –1 1 1 1 Similarly, the direction cosines of AF, BG and CD are , , ; , 3333 –1 1 1 1 –1 , and , , , respectively. 33 3 33 Let l, m, n be the direction cosines of the given line which makes angles α, β, γ, δ with OE, AF, BG, CD, respectively. Then cosα = 11 (l + m+ n); cos β = 3 (– l + m + n); 3 1 1 (Why?) cosγ = (l – m + n); cos δ = 3 (l + m – n) 3 Squaring and adding, we get cos2α + cos2 β + cos2 γ + cos2 δ 1 = [ (l + m + n )2 + (–l + m + n)2 ] + (l – m + n)2 + (l + m –n)2] 3 14 (as l2 + m2 + n2 = 1) = [ 4 (l2 + m2 + n2 ) ] = 33 Example 27 Find the equation of the plane that contains the point (1, – 1, 2) and is perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Solution The equation of the plane containing the given point is ... (1) A (x – 1) + B(y + 1) + C (z – 2) = 0 Applying the condition of perpendicularly to the plane given in (1) with the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8, we have 2A + 3B – 2C = 0 and A + 2B – 3C = 0 Solving these equations, we find A = – 5C and B = 4C. Hence, the required equation is – 5C (x – 1) + 4 C (y + 1) + C(z – 2) = 0 i.e. 5x – 4y – z = 7 Example 28 Find the distance between the point P(6, 5, 9) and the plane determined by the points A (3, – 1, 2), B (5, 2, 4) and C(– 1, – 1, 6). Solution Let A, B, C be the three points in the plane. D is the foot of the perpendicular drawn from a point Pto the plane. PDis the required distance to be determined, which is the projection of AP on AB × AC . 2019-20

496 MATHEMATICS    Hence, PD = the dot product of AP with the unit vector along AB × AC .  So AP = 3 iˆ + 6 ˆj + 7 kˆ and   iˆ ˆj kˆ AB × AC = 2 3 2 = 12iˆ − 16 ˆj + 12 kˆ −4 0 4   3 iˆ − 4 ˆj + 3 kˆ Unit vector along AB × AC = 34 Hence PD = ( 3 iˆ + 6 ˆj + 7 kˆ ) . 3 iˆ − 4 ˆj + 3 kˆ 34 3 34 = 17 Alternatively, find the equation of the plane passing through A, B and C and then compute the distance of the point P from the plane. Example 29 Show that the lines x−a+d = y−a = z−a−d α−δ α α+δ and x−b+c = y−b = z −b −c are coplanar. β−γ β β+γ Solution Here x1 = a – d x2 = b – c y=b y=a 1 2 z1 = a + d z2 = b + c a2 = β – γ a1 = α – δ b2 = β c2 = β + γ b1 = α c1 = α + δ Now consider the determinant x2 − x1 y2 − y1 z2 − z1 b − c − a + d b−a b +c − a − d a1 b1 α α+δ a2 b2 c1 = α−δ β β+γ c2 β−γ 2019-20

THREE DIMENSIONAL GEOMETRY 497 Adding third column to the first column, we get b − a b − a b +c − a − d 2α α α +δ =0 ββ β+γ Since the first and second columns are identical. Hence, the given two lines are coplanar. Example 30 Find the coordinates of the point where the line through the points A (3, 4, 1) and B (5, 1, 6) crosses the XY-plane. Solution The vector equation of the line through the points A and B is ... (1) r = 3 iˆ + 4 ˆj + kˆ + λ [ (5 −3)iˆ + (1 − 4) ˆj + (6 − 1) kˆ ] i.e. r = 3 iˆ + 4 ˆj + kˆ + λ ( 2 iˆ −3 ˆj + 5 kˆ ) Let P be the point where the line AB crosses the XY-plane. Then the position vector of the point P is of the form x iˆ + y ˆj . This point must satisfy the equation (1). (Why ?) i.e. x iˆ + y ˆj = (3 + 2 λ ) iˆ + ( 4 − 3 λ) ˆj + ( 1 + 5 λ ) kˆ Equating the like coefficients of iˆ, ˆj and kˆ , we have x= 3+2λ y= 4–3λ 0= 1+5λ Solving the above equations, we get x = 13 and y = 23 55 Hence, the coordinates of the required point are 13 , 23 , 0  . 5 5  Miscellaneous Exercise on Chapter 11 1. Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, – 1), (4, 3, – 1). 2. If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m1 n2 − m2 n1 , n1 l2 − n2 l1 , l1 m2 − l2 m1 2019-20

498 MATHEMATICS 3. Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b. 4. Find the equation of a line parallel to x-axis and passing through the origin. 5. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD. 6. If the lines x − 1 = y − 2 = z −3 and x −1 = y − 1 = z − 6 are perpendicular, −3 2k 2 3k 1 −5 find the value of k. 7. Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane r . ( iˆ + 2 ˆj − 5 kˆ ) + 9 = 0 . 8. Find the equation of the plane passing through (a, b, c) and parallel to the plane r ⋅ (iˆ + ˆj + kˆ) = 2. 9. Find the shortest distance between lines r = 6 iˆ + 2 ˆj + 2 kˆ + λ (iˆ − 2 ˆj + 2 kˆ) and r = − 4 iˆ − kˆ + µ (3 iˆ − 2 ˆj − 2 kˆ) . 10. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ-plane. 11. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane. 12. Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane 2x + y + z = 7. 13. Find the equation of the plane passing through the point (– 1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0. 14. If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane r ⋅ (3 iˆ + 4 ˆj − 12 kˆ) + 13 = 0, then find the value of p. 15. Find the equation of the plane passing through the line of intersection of the planes r ⋅ (iˆ + ˆj + kˆ) = 1 and r ⋅ (2 iˆ + 3 ˆj − kˆ) + 4 = 0 and parallel to x-axis. 16. If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing through P and perpendicular to OP. 17. Find the equation of the plane which contains the line of intersection of the planes r ⋅ (iˆ + 2 ˆj + 3 kˆ) − 4 = 0 , r ⋅ (2 iˆ + ˆj − kˆ) + 5 = 0 and which is perpendicular to the plane r ⋅ (5 iˆ + 3 ˆj − 6 kˆ) + 8 = 0 . 2019-20

THREE DIMENSIONAL GEOMETRY 499 18. Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line r = 2 iˆ − ˆj + 2 kˆ + λ (3 iˆ + 4 ˆj + 2 kˆ) and the plane r ⋅ (iˆ − ˆj + kˆ) = 5 . 19. Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes r ⋅ (iˆ − ˆj + 2 kˆ) = 5 and r ⋅ (3 iˆ + ˆj + kˆ) = 6 . 20. Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines: x−8 = y + 19 = z −10 and x − 15 = y − 29 = z− 5 . 3 −16 7 3 8 −5 21. Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then 1 + 1 + 1 = 1 . a2 b2 c2 p2 Choose the correct answer in Exercises 22 and 23. 22. Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is (A) 2 units (B) 4 units (C) 8 units (D) 2 units 29 23. The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are (A) Perpendicular (B) Parallel (C) intersect y-axis (D) passes through  0, 0, 5   4  Summary  Direction cosines of a line are the cosines of the angles made by the line with the positive directions of the coordinate axes.  If l, m, n are the direction cosines of a line, then l2 + m2 + n2 = 1.  Direction cosines of a line joining two points P(x1, y1, z1) and Q(x2, y2, z2) are x2 − x1 , y2 − y1 , z2 − z1 PQ PQ PQ where PQ = ( )(x2 − x1)2 + ( y2 − y1)2 + z2 − z1 2  Direction ratios of a line are the numbers which are proportional to the direction cosines of a line.  If l, m, n are the direction cosines and a, b, c are the direction ratios of a line 2019-20

500 MATHEMATICS then abc l= ;m= ;n= a2 + b2 + c2 a2 + b2 + c2 a2 + b2 + c2  Skew lines are lines in space which are neither parallel nor intersecting. They lie in different planes.  Angle between skew lines is the angle between two intersecting lines drawn from any point (preferably through the origin) parallel to each of the skew lines.  If l1, m1, n1 and l2, m2, n2 are the direction cosines of two lines; and θ is the acute angle between the two lines; then cosθ = | l1l2 + m1m2 + n1n2|  If a1, b1, c1 and a2, b2, c2 are the direction ratios of two lines and θ is the acute angle between the two lines; then cosθ = a1 a2 + b1 b2 + c1 c2 a12 + b12 + c12 a22 + b22 + c22  Vector equation of a line that passes through the given point whose position vector is a and parallel to a given vector b is r = a +λb .  Equation of a line through a point (x , y , z ) and having direction cosines l, m, n is 1 11 x − x1 = y − y1 = z − z1 l mn  Tvehcetovresctaorreeqauaatnidonbofias line which p(bass−esat)h.rough two points whose position r = a + λ  Cartesian equation of a line that passes through two points (x1, y1, z1) and  (x , y , z ) is x − x1 = y − y1 = z − z1 . λ  and r = a2  then 22 2 x2 − x1 y2 − y1 en z2 − z1 + b1 + λ b2 , the acute If θ is a ngle betwe r = a1  cosθ = b1 ⋅ b2 | b1 | | b2 |  If x − x1 = y − y1 = z − z1 and x − x2 = y − y 2 = z − z 2 l1 m1 n1 l2 m2 n2 are the equations of two lines, then the acute angle between the two lines is given by cos θ = |l1l2 + m1 m2 + n1n2|. 2019-20

THREE DIMENSIONAL GEOMETRY 501  Shortest distance between two skew lines is the line segment perpendicular to both the lines.    Shortest distance between r = a1 + λ b1 and r = a2 + µ b2 is  ×  ) ⋅ (a2 – a1 ) (b1 b2 | b1 × b2 |  Shortest distance between the lines: x − x1 = y − y1 = z − z1 and a1 b1 c1 x − x2 = y − y2 = z − z2 is a2 b2 c2 x2 − x1 y2 − y1 z2 − z1 a1 b1 c1 a2 b2 c2 (b1c2 − b2c1)2 + (c1a2 − c2a1)2 + (a1b2 − a2b1)2  Distance between parallel lines r = a1 +  and r = a2 + µ  is λb b  × (a2 −  ) b a1 |b |  In the vector form, equation of a plane which is at a distance d from the orr⋅ignˆin=, and nˆ is the unit vector normal to the plane through the origin is d.  Equation of a plane which is at a distance of d from the origin and the direction cosines of the normal to the plane as l, m, n is lx + my + nz = d. a  The equation of taheplvaencetotrhrNouigsh(arp−oiant) whose position vector is and perpendicular to . N = 0.  Equation of a plane perpendicular to a given line with direction ratios A, B, C and passing through a given point (x1, y1, z1) is A (x – x1) + B (y – y1) + C (z – z1 ) = 0  Equation of a plane passing through three non collinear points (x1, y1, z1), 2019-20

502 MATHEMATICS (x2, y2, z2) and (x3, y3, z3) is y − y1 z − z1 =0 y2 − y1 z2 − z1 x − x1 y3 − y1 z3 − z1 x2 − x1 x3 − x1  Vector equation of a plane that contains three non collinear points having position vectors a, b and  ( r − a ) . [(b − a) × ( c − a )]=0 c is  Equation of a plane that cuts the coordinates axes at (a, 0, 0), (0, b, 0) and (0, 0, c) is x + y+z =1 a bc  Vector equation of a plane that passes through the intersection of planes r ⋅ n1 = d1 and r ⋅ n2 = d2 is r ⋅ (n1 + λ n2 ) = d1 + λ d2 , where λ is any nonzero constant.  Cartesian equation of a plane that passes through the intersection of two given planes A1 x + B1 y + C1 z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0 is (A1 x + B1 y + C1 z + Dan1()da+2r−λ=(aA1)a2⋅2x(b1++×µBb22b)y2 + C2 z + D2) = 0. Two lines r = a1 + λ  are coplanar if  b1 =0  In the cartesian form two lines = x − x1 = y − y1 = z − z1 and x − x2 a1 b1 c1 a2 = y – y2 = z – z2 are coplanar if x2 − x1 y2 − y1 z2 − z1 b2 C2 a1 b1 a2 b2 c1 = 0. c2  In the vector form, if θ is the angle between the two planes, r ⋅ n1 = d1 and  ,φthbeentwθee=nctohse–1lin|| nen11|r⋅| nn=22 the plane r ⋅ nˆ =d is r ⋅ n2 = d2 | λ  and The angle . b |a + 2019-20

THREE DIMENSIONAL GEOMETRY 503  sin φ = b ⋅ nˆ |b | |nˆ|  The angle θ between the planes A1x + B1y + C1z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0 is given by cos θ = A1 A2 + B1 B2 + C1 C2 A12 + B12 + C12 A22 + B22 + C22  T| dhe−dais⋅tnaˆn| ce of a point whose position vector is a from the plane r ⋅ nˆ = d is  The distance from a point (x1, y1, z1) to the plane Ax + By + Cz + D = 0 is Ax1 + By1 + Cz1 + D . A2 + B2 + C2 —— 2019-20

504 MATHEMATICS 12Chapter LINEAR PROGRAMMING  The mathematical experience of the student is incomplete if he never had the opportunity to solve a problem invented by himself. – G. POLYA  12.1 Introduction In earlier classes, we have discussed systems of linear equations and their applications in day to day problems. In Class XI, we have studied linear inequalities and systems of linear inequalities in two variables and their solutions by graphical method. Many applications in mathematics involve systems of inequalities/equations. In this chapter, we shall apply the systems of linear inequalities/equations to solve some real life problems of the type as given below: A furniture dealer deals in only two items–tables and chairs. He has Rs 50,000 to invest and has storage space of at most 60 pieces. A table costs Rs 2500 and a chair Rs 500. He estimates that from the sale of one table, he L. Kantorovich can make a profit of Rs 250 and that from the sale of one chair a profit of Rs 75. He wants to know how many tables and chairs he should buy from the available money so as to maximise his total profit, assuming that he can sell all the items which he buys. Such type of problems which seek to maximise (or, minimise) profit (or, cost) form a general class of problems called optimisation problems. Thus, an optimisation problem may involve finding maximum profit, minimum cost, or minimum use of resources etc. A special but a very important class of optimisation problems is linear programming problem. The above stated optimisation problem is an example of linear programming problem. Linear programming problems are of much interest because of their wide applicability in industry, commerce, management science etc. In this chapter, we shall study some linear programming problems and their solutions by graphical method only, though there are many other methods also to solve such problems. 2019-20

LINEAR PROGRAMMING 505 12.2 Linear Programming Problem and its Mathematical Formulation We begin our discussion with the above example of furniture dealer which will further lead to a mathematical formulation of the problem in two variables. In this example, we observe (i) The dealer can invest his money in buying tables or chairs or combination thereof. Further he would earn different profits by following different investment strategies. (ii) There are certain overriding conditions or constraints viz., his investment is limited to a maximum of Rs 50,000 and so is his storage space which is for a maximum of 60 pieces. Suppose he decides to buy tables only and no chairs, so he can buy 50000 ÷ 2500, i.e., 20 tables. His profit in this case will be Rs (250 × 20), i.e., Rs 5000. Suppose he chooses to buy chairs only and no tables. With his capital of Rs 50,000, he can buy 50000 ÷ 500, i.e. 100 chairs. But he can store only 60 pieces. Therefore, he is forced to buy only 60 chairs which will give him a total profit of Rs (60 × 75), i.e., Rs 4500. There are many other possibilities, for instance, he may choose to buy 10 tables and 50 chairs, as he can store only 60 pieces. Total profit in this case would be Rs (10 × 250 + 50 × 75), i.e., Rs 6250 and so on. We, thus, find that the dealer can invest his money in different ways and he would earn different profits by following different investment strategies. Now the problem is : How should he invest his money in order to get maximum profit? To answer this question, let us try to formulate the problem mathematically. 12.2.1 Mathematical formulation of the problem Let x be the number of tables and y be the number of chairs that the dealer buys. Obviously, x and y must be non-negative, i.e., x≥0 ... (1) (Non-negative constraints) ... (2) y≥0 The dealer is constrained by the maximum amount he can invest (Here it is Rs 50,000) and by the maximum number of items he can store (Here it is 60). Stated mathematically, 2500x + 500y ≤ 50000 (investment constraint) or 5x + y ≤ 100 ... (3) and x + y ≤ 60 (storage constraint) ... (4) 2019-20

506 MATHEMATICS The dealer wants to invest in such a way so as to maximise his profit, say, Z which stated as a function of x and y is given by Z = 250x + 75y (called objective function) ... (5) Mathematically, the given problems now reduces to: Maximise Z = 250x + 75y subject to the constraints: 5x + y ≤ 100 x + y ≤ 60 x ≥ 0, y ≥ 0 So, we have to maximise the linear function Z subject to certain conditions determined by a set of linear inequalities with variables as non-negative. There are also some other problems where we have to minimise a linear function subject to certain conditions determined by a set of linear inequalities with variables as non-negative. Such problems are called Linear Programming Problems. Thus, a Linear Programming Problem is one that is concerned with finding the optimal value (maximum or minimum value) of a linear function (called objective function) of several variables (say x and y), subject to the conditions that the variables are non-negative and satisfy a set of linear inequalities (called linear constraints). The term linear implies that all the mathematical relations used in the problem are linear relations while the term programming refers to the method of determining a particular programme or plan of action. Before we proceed further, we now formally define some terms (which have been used above) which we shall be using in the linear programming problems: Objective function Linear function Z = ax + by, where a, b are constants, which has to be maximised or minimized is called a linear objective function. In the above example, Z = 250x + 75y is a linear objective function. Variables x and y are called decision variables. Constraints The linear inequalities or equations or restrictions on the variables of a linear programming problem are called constraints. The conditions x ≥ 0, y ≥ 0 are called non-negative restrictions. In the above example, the set of inequalities (1) to (4) are constraints. Optimisation problem A problem which seeks to maximise or minimise a linear function (say of two variables x and y) subject to certain constraints as determined by a set of linear inequalities is called an optimisation problem. Linear programming problems are special type of optimisation problems. The above problem of investing a 2019-20

LINEAR PROGRAMMING 507 given sum by the dealer in purchasing chairs and tables is an example of an optimisation problem as well as of a linear programming problem. We will now discuss how to find solutions to a linear programming problem. In this chapter, we will be concerned only with the graphical method. 12.2.2 Graphical method of solving linear programming problems In Class XI, we have learnt how to graph a system of linear inequalities involving two variables x and y and to find its solutions graphically. Let us refer to the problem of investment in tables and chairs discussed in Section 12.2. We will now solve this problem graphically. Let us graph the constraints stated as linear inequalities: 5x + y ≤ 100 ... (1) x + y ≤ 60 ... (2) x≥0 ... (3) y≥0 ... (4) The graph of this system (shaded region) consists of the points common to all half planes determined by the inequalities (1) to (4) (Fig 12.1). Each point in this region represents a feasible choice open to the dealer for investing in tables and chairs. The region, therefore, is called the feasible region for the problem. Every point of this region is called a feasible solution to the problem. Thus, we have, Feasible region The common region determined by all the constraints including non-negative constraints x, y ≥ 0 of a linear programming problem is called the feasible region (or solution region) for the problem. In Fig 12.1, the region OABC (shaded) is the feasible region for the problem. The region other than feasible region is called an infeasible region. Feasible solutions Points within and on the boundary of the feasible region represent feasible solutions of the constraints. In Fig 12.1, every point within and on the boundary of the feasible region OABC represents feasible solution to the problem. For example, the point (10, 50) is a feasible solution of the problem and so are the points (0, 60), (20, 0) etc. Any point outside the feasible region is Fig 12.1 called an infeasible solution. For example, the point (25, 40) is an infeasible solution of the problem. 2019-20

508 MATHEMATICS Optimal (feasible) solution: Any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an optimal solution. Now, we see that every point in the feasible region OABC satisfies all the constraints as given in (1) to (4), and since there are infinitely many points, it is not evident how we should go about finding a point that gives a maximum value of the objective function Z = 250x + 75y. To handle this situation, we use the following theorems which are fundamental in solving linear programming problems. The proofs of these theorems are beyond the scope of the book. Theorem 1 Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities, this optimal value must occur at a corner point* (vertex) of the feasible region. Theorem 2 Let R be the feasible region for a linear programming problem, and let Z = ax + by be the objective function. If R is bounded**, then the objective function Z has both a maximum and a minimum value on R and each of these occurs at a corner point (vertex) of R. Remark If R is unbounded, then a maximum or a minimum value of the objective function may not exist. However, if it exists, it must occur at a corner point of R. (By Theorem 1). In the above example, the corner points (vertices) of the bounded (feasible) region are: O, A, B and C and it is easy to find their coordinates as (0, 0), (20, 0), (10, 50) and (0, 60) respectively. Let us now compute the values of Z at these points. We have Vertex of the Corresponding value Feasible Region of Z (in Rs) O (0,0) 0 ← Maximum C (0,60) 4500 B (10,50) 6250 A (20,0) 5000 * A corner point of a feasible region is a point in the region which is the intersection of two boundary lines. ** A feasible region of a system of linear inequalities is said to be bounded if it can be enclosed within a circle. Otherwise, it is called unbounded. Unbounded means that the feasible region does extend indefinitely in any direction. 2019-20

LINEAR PROGRAMMING 509 We observe that the maximum profit to the dealer results from the investment strategy (10, 50), i.e. buying 10 tables and 50 chairs. This method of solving linear programming problem is referred as Corner Point Method. The method comprises of the following steps: 1. Find the feasible region of the linear programming problem and determine its corner points (vertices) either by inspection or by solving the two equations of the lines intersecting at that point. 2. Evaluate the objective function Z = ax + by at each corner point. Let M and m, respectively denote the largest and smallest values of these points. 3. (i) When the feasible region is bounded, M and m are the maximum and minimum values of Z. (ii) In case, the feasible region is unbounded, we have: 4. (a) M is the maximum value of Z, if the open half plane determined by ax + by > M has no point in common with the feasible region. Otherwise, Z has no maximum value. (b) Similarly, m is the minimum value of Z, if the open half plane determined by ax + by < m has no point in common with the feasible region. Otherwise, Z has no minimum value. We will now illustrate these steps of Corner Point Method by considering some examples: Example 1 Solve the following linear programming problem graphically: Maximise Z = 4x + y ... (1) subject to the constraints: x + y ≤ 50 ... (2) 3x + y ≤ 90 ... (3) x ≥ 0, y ≥ 0 ... (4) Solution The shaded region in Fig 12.2 is the feasible region determined by the system of constraints (2) to (4). We observe that the feasible region OABC is bounded. So, we now use Corner Point Method to determine the maximum value of Z. The coordinates of the corner points O, A, B and C are (0, 0), (30, 0), (20, 30) and (0, 50) respectively. Now we evaluate Z at each corner point. 2019-20

510 MATHEMATICS Corner Point Corresponding value of Z (0, 0) 0 Maximum (30, 0) (20, 30) 120 ← (0, 50) 110 50 Fig 12.2 ... (1) Hence, maximum value of Z is 120 at the point (30, 0). ... (2) ... (3) Example 2 Solve the following linear programming problem graphically: ... (4) Minimise Z = 200 x + 500 y subject to the constraints: x + 2y ≥ 10 3x + 4y ≤ 24 x ≥ 0, y ≥ 0 Solution The shaded region in Fig 12.3 is the feasible region ABC determined by the system of constraints (2) to (4), which is bounded. The coordinates of corner points Corner Point Corresponding value of Z (0, 5) 2500 (4, 3) (0, 6) ←2300 Minimum 3000 Fig 12.3 2019-20

LINEAR PROGRAMMING 511 A, B and C are (0,5), (4,3) and (0,6) respectively. Now we evaluate Z = 200x + 500y at these points. Hence, minimum value of Z is 2300 attained at the point (4, 3) Example 3 Solve the following problem graphically: ... (1) Minimise and Maximise Z = 3x + 9y ... (2) subject to the constraints: x + 3y ≤ 60 ... (3) ... (4) x + y ≥ 10 ... (5) x ≤y x ≥ 0, y ≥ 0 Solution First of all, let us graph the feasible region of the system of linear inequalities (2) to (5). The feasible region ABCD is shown in the Fig 12.4. Note that the region is bounded. The coordinates of the corner points A, B, C and D are (0, 10), (5, 5), (15,15) and (0, 20) respectively. Corner Corresponding value of Point Z = 3x + 9y A (0, 10) 90 Minimum B (5, 5) Maximum C (15, 15) 60 ← (Multiple D (0, 20) optimal }180 ← solutions) 180 Fig 12.4 We now find the minimum and maximum value of Z. From the table, we find that the minimum value of Z is 60 at the point B (5, 5) of the feasible region. The maximum value of Z on the feasible region occurs at the two corner points C (15, 15) and D (0, 20) and it is 180 in each case. Remark Observe that in the above example, the problem has multiple optimal solutions at the corner points C and D, i.e. the both points produce same maximum value 180. In such cases, you can see that every point on the line segment CD joining the two corner points C and D also give the same maximum value. Same is also true in the case if the two points produce same minimum value. 2019-20

512 MATHEMATICS Example 4 Determine graphically the minimum value of the objective function Z = – 50x + 20y ... (1) subject to the constraints: 2x – y ≥ – 5 ... (2) 3x + y ≥ 3 ... (3) 2x – 3y ≤ 12 ... (4) x ≥ 0, y ≥ 0 ... (5) Solution First of all, let us graph the feasible region of the system of inequalities (2) to (5). The feasible region (shaded) is shown in the Fig 12.5. Observe that the feasible region is unbounded. We now evaluate Z at the corner points. Corner Point Z = – 50x + 20y (0, 5) 100 (0, 3) (1, 0) 60 (6, 0) –50 ←– 300 smallest Fig 12.5 From this table, we find that – 300 is the smallest value of Z at the corner point (6, 0). Can we say that minimum value of Z is – 300? Note that if the region would have been bounded, this smallest value of Z is the minimum value of Z (Theorem 2). But here we see that the feasible region is unbounded. Therefore, – 300 may or may not be the minimum value of Z. To decide this issue, we graph the inequality – 50x + 20y < – 300 (see Step 3(ii) of corner Point Method.) i.e., – 5x + 2y < – 30 and check whether the resulting open half plane has points in common with feasible region or not. If it has common points, then –300 will not be the minimum value of Z. Otherwise, –300 will be the minimum value of Z. 2019-20

LINEAR PROGRAMMING 513 As shown in the Fig 12.5, it has common points. Therefore, Z = –50 x + 20 y has no minimum value subject to the given constraints. In the above example, can you say whether z = – 50 x + 20 y has the maximum value 100 at (0,5)? For this, check whether the graph of – 50 x + 20 y > 100 has points in common with the feasible region. (Why?) Example 5 Minimise Z = 3x + 2y ... (1) subject to the constraints: ... (2) ... (3) x+y≥8 3x + 5y ≤ 15 x ≥ 0, y ≥ 0 Solution Let us graph the inequalities (1) to (3) (Fig 12.6). Is there any feasible region? Why is so? From Fig 12.6, you can see that there is no point satisfying all the constraints simultaneously. Thus, the problem is having no feasible region and hence no feasible solution. Remarks From the examples which we have discussed so far, we notice some general features of linear programming problems: (i) The feasible region is always a convex region. Fig 12.6 (ii) The maximum (or minimum) solution of the objective function occurs at the vertex (corner) of the feasible region. If two corner points produce the same maximum (or minimum) value of the objective function, then every point on the line segment joining these points will also give the same maximum (or minimum) value. EXERCISE 12.1 Solve the following Linear Programming Problems graphically: 1. Maximise Z = 3x + 4y subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0. 2019-20

514 MATHEMATICS 2. Minimise Z = – 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0. 3. Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0. 4. Minimise Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0. 5. Maximise Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0. 6. Minimise Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0. Show that the minimum of Z occurs at more than two points. 7. Minimise and Maximise Z = 5x + 10 y subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0. 8. Minimise and Maximise Z = x + 2y subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0. 9. Maximise Z = – x + 2y, subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0. 10. Maximise Z = x + y, subject to x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0. 12.3 Different Types of Linear Programming Problems A few important linear programming problems are listed below: 1. Manufacturing problems In these problems, we determine the number of units of different products which should be produced and sold by a firm when each product requires a fixed manpower, machine hours, labour hour per unit of product, warehouse space per unit of the output etc., in order to make maximum profit. 2. Diet problems In these problems, we determine the amount of different kinds of constituents/nutrients which should be included in a diet so as to minimise the cost of the desired diet such that it contains a certain minimum amount of each constituent/nutrients. 3. Transportation problems In these problems, we determine a transportation schedule in order to find the cheapest way of transporting a product from plants/factories situated at different locations to different markets. 2019-20

LINEAR PROGRAMMING 515 Let us now solve some of these types of linear programming problems: Example 6 (Diet problem): A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast 8 units of vitamin A and 10 units of vitamin C. Food ‘I’ contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C. Food ‘II’ contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs Rs 50 per kg to purchase Food ‘I’ and Rs 70 per kg to purchase Food ‘II’. Formulate this problem as a linear programming problem to minimise the cost of such a mixture. Solution Let the mixture contain x kg of Food ‘I’ and y kg of Food ‘II’. Clearly, x ≥ 0, y ≥ 0. We make the following table from the given data: Resources Food Requirement I II Vitamin A (x) (y) 8 (units/kg) 21 10 Vitamin C (units/kg) 12 Cost (Rs/kg) 50 70 Since the mixture must contain at least 8 units of vitamin A and 10 units of vitamin C, we have the constraints: 2x + y ≥ 8 x + 2y ≥ 10 Total cost Z of purchasing x kg of food ‘I’ and y kg of Food ‘II’ is Z = 50x + 70y Hence, the mathematical formulation of the problem is: Minimise Z = 50x + 70y ... (1) subject to the constraints: 2x + y ≥ 8 ... (2) x + 2y ≥ 10 ... (3) x, y ≥ 0 ... (4) Let us graph the inequalities (2) to (4). The feasible region determined by the system is shown in the Fig 12.7. Here again, observe that the feasible region is unbounded. 2019-20

516 MATHEMATICS Let us evaluate Z at the corner points A(0,8), B(2,4) and C(10,0). Corner Point Z = 50x + 70y (0,8) 560 (2,4) (10,0) ←380 Minimum 500 Fig 12.7 In the table, we find that smallest value of Z is 380 at the point (2,4). Can we say that the minimum value of Z is 380? Remember that the feasible region is unbounded. Therefore, we have to draw the graph of the inequality 50x + 70y < 380 i.e., 5x + 7y < 38 to check whether the resulting open half plane has any point common with the feasible region. From the Fig 12.7, we see that it has no points in common. Thus, the minimum value of Z is 380 attained at the point (2, 4). Hence, the optimal mixing strategy for the dietician would be to mix 2 kg of Food ‘I’ and 4 kg of Food ‘II’, and with this strategy, the minimum cost of the mixture will be Rs 380. Example 7 (Allocation problem) A cooperative society of farmers has 50 hectare of land to grow two crops X and Y. The profit from crops X and Y per hectare are estimated as Rs 10,500 and Rs 9,000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y at rates of 20 litres and 10 litres per hectare. Further, no more than 800 litres of herbicide should be used in order to protect fish and wild life using a pond which collects drainage from this land. How much land should be allocated to each crop so as to maximise the total profit of the society? Solution Let x hectare of land be allocated to crop X and y hectare to crop Y. Obviously, x ≥ 0, y ≥ 0. Profit per hectare on crop X = Rs 10500 Profit per hectare on crop Y = Rs 9000 Therefore, total profit = Rs (10500x + 9000y) 2019-20

LINEAR PROGRAMMING 517 The mathematical formulation of the problem is as follows: Maximise Z = 10500 x + 9000 y subject to the constraints: x + y ≤ 50 (constraint related to land) ... (1) 20x + 10y ≤ 800 (constraint related to use of herbicide) i.e. 2x + y ≤ 80 ... (2) x ≥ 0, y ≥ 0 (non negative constraint) ... (3) Let us draw the graph of the system of inequalities (1) to (3). The feasible region OABC is shown (shaded) in the Fig 12.8. Observe that the feasible region is bounded. The coordinates of the corner points O, A, B and C are (0, 0), (40, 0), (30, 20) and (0, 50) respectively. Let us evaluate the objective function Z = 10500 x + 9000y at these vertices to find which one gives the maximum profit. Corner Point Z = 10500x + 9000y O (0, 0) 0 A ( 40, 0) 420000 B (30, 20) 495000 ← Maximum C (0,50) 450000 Fig 12.8 Hence, the society will get the maximum profit of Rs 4,95,000 by allocating 30 hectares for crop X and 20 hectares for crop Y. Example 8 (Manufacturing problem) A manufacturing company makes two models A and B of a product. Each piece of Model A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each piece of Model B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 180 and 30 respectively. The company makes a profit of Rs 8000 on each piece of model A and Rs 12000 on each piece of Model B. How many pieces of Model A and Model B should be manufactured per week to realise a maximum profit? What is the maximum profit per week? 2019-20

518 MATHEMATICS Solution Suppose x is the number of pieces of Model A and y is the number of pieces of Model B. Then Total profit (in Rs) = 8000 x + 12000 y Let Z = 8000 x + 12000 y We now have the following mathematical model for the given problem. Maximise Z = 8000 x + 12000 y ... (1) subject to the constraints: 9x + 12y ≤ 180 (Fabricating constraint) i.e. 3x + 4y ≤ 60 ... (2) x + 3y ≤ 30 (Finishing constraint) ... (3) x ≥ 0, y ≥ 0 (non-negative constraint) ... (4) The feasible region (shaded) OABC determined by the linear inequalities (2) to (4) is shown in the Fig 12.9. Note that the feasible region is bounded. Fig 12.9 Let us evaluate the objective function Z at each corner point as shown below: Corner Point Z = 8000 x + 12000 y 0 (0, 0) 0 A (20, 0) 160000 B (12, 6) ←168000 Maximum C (0, 10) 120000 We find that maximum value of Z is 1,68,000 at B (12, 6). Hence, the company should produce 12 pieces of Model A and 6 pieces of Model B to realise maximum profit and maximum profit then will be Rs 1,68,000. 2019-20

LINEAR PROGRAMMING 519 EXERCISE 12.2 1. Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs 80/kg. Food P contains 3 units/kg of Vitamin A and 5 units / kg of Vitamin B while food Q contains 4 units/kg of Vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture. 2. One kind of cake requires 200g of flour and 25g of fat, and another kind of cake requires 100g of flour and 50g of fat. Find the maximum number of cakes which can be made from 5kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes. 3. A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time. (i) What number of rackets and bats must be made if the factory is to work at full capacity? (ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity. 4. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs17.50 per package on nuts and Rs 7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day? 5. A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit. 2019-20

520 MATHEMATICS 6. A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs 5 and that from a shade is Rs 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit? 7. A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit? 8. A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000. 9. A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F and F are available. Food F costs Rs 4 per unit food and F costs 12 1 2 Rs 6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F contains 6 units of vitamin A and 3 units of minerals. 2 Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements. 10. There are two types of fertilisers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs Rs 6/kg and F2 costs Rs 5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost? 11. The corner points of the feasible region determined by the following system of linear inequalities: 2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is (A) p = q (B) p = 2q (C) p = 3q (D) q = 3p 2019-20

LINEAR PROGRAMMING 521 Miscellaneous Examples Example 9 (Diet problem) A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol. How many packets of each food should be used to minimise the amount of vitamin A in the diet? What is the minimum amount of vitaminA? Solution Let x and y be the number of packets of food P and Q respectively. Obviously x ≥ 0, y ≥ 0. Mathematical formulation of the given problem is as follows: Minimise Z = 6x + 3y (vitamin A) subject to the constraints 12x + 3y ≥ 240 (constraint on calcium), i.e. 4x + y ≥ 80 ... (1) 4x + 20y ≥ 460 (constraint on iron), i.e. x + 5y ≥ 115 ... (2) 6x + 4y ≤ 300 (constraint on cholesterol), i.e. 3x + 2y ≤ 150 ... (3) x ≥ 0, y ≥ 0 ... (4) Let us graph the inequalities (1) to (4). The feasible region (shaded) determined by the constraints (1) to (4) is shown in Fig 12.10 and note that it is bounded. Fig 12.10 2019-20

522 MATHEMATICS The coordinates of the corner points L, M and N are (2, 72), (15, 20) and (40, 15) respectively. Let us evaluate Z at these points: Corner Point Z=6x+3y Minimum (2, 72) 228 (15, 20) (40, 15) 150 ← 285 From the table, we find that Z is minimum at the point (15, 20). Hence, the amount of vitamin A under the constraints given in the problem will be minimum, if 15 packets of food P and 20 packets of food Q are used in the special diet. The minimum amount of vitamin A will be 150 units. Example 10 (Manufacturing problem) A manufacturer has three machines I, II and III installed in his factory. Machines I and II are capable of being operated for at most 12 hours whereas machine III must be operated for atleast 5 hours a day. She produces only two items M and N each requiring the use of all the three machines. The number of hours required for producing 1 unit of each of M and N on the three machines are given in the following table: Items Number of hours required on machines I II III M1 21 N2 1 1.25 She makes a profit of Rs 600 and Rs 400 on items M and N respectively. How many of each item should she produce so as to maximise her profit assuming that she can sell all the items that she produced? What will be the maximum profit? Solution Let x and y be the number of items M and N respectively. ... (1) Total profit on the production = Rs (600 x + 400 y) ... (2) Mathematical formulation of the given problem is as follows: Maximise Z = 600 x + 400 y subject to the constraints: x + 2y ≤ 12 (constraint on Machine I) 2x + y ≤ 12 (constraint on Machine II) 5 ... (3) x + 4 y ≥ 5 (constraint on Machine III) ... (4) x ≥ 0, y ≥ 0 2019-20

LINEAR PROGRAMMING 523 Let us draw the graph of constraints (1) to (4). ABCDE is the feasible region (shaded) as shown in Fig 12.11 determined by the constraints (1) to (4). Observe that the feasible region is bounded, coordinates of the corner points A, B, C, D and E are (5, 0) (6, 0), (4, 4), (0, 6) and (0, 4) respectively. Fig 12.11 Let us evaluate Z = 600 x + 400 y at these corner points. Corner point Z = 600 x + 400 y Maximum (5, 0) 3000 (6, 0) 3600 (4, 4) (0, 6) ←4000 (0, 4) 2400 1600 We see that the point (4, 4) is giving the maximum value of Z. Hence, the manufacturer has to produce 4 units of each item to get the maximum profit of Rs 4000. Example 11 (Transportation problem) There are two factories located one at place P and the other at place Q. From these locations, a certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are respectively 8 and 6 units. The cost of 2019-20

524 MATHEMATICS transportation per unit is given below: Cost (in Rs) From/To ABC P 160 100 150 Q 100 120 100 How many units should be transported from each factory to each depot in order that the transportation cost is minimum. What will be the minimum transportation cost? Solution The problem can be explained diagrammatically as follows (Fig 12.12): Let x units and y units of the commodity be transported from the factory at P to the depots at A and B respectively. Then (8 – x – y) units will be transported to depot at C (Why?) Factory P 8 units x 160 8– xRs 150 Rs y – y Rs 100 A B C Depot 5 units 5 units Depot 4 units Depot Rs 100 Rs 100 – y)] x 5 – 5 – y Rs 120 + (5 6 – x) Q 6– [(5 units Factory Fig 12.12 Hence, we have x ≥ 0, y ≥ 0 and 8 – x – y ≥ 0 i.e. x ≥ 0, y ≥ 0 and x + y ≤ 8 Now, the weekly requirement of the depot at A is 5 units of the commodity. Since x units are transported from the factory at P, the remaining (5 – x) units need to be transported from the factory at Q. Obviously, 5 – x ≥ 0, i.e. x ≤ 5. Similarly, (5 – y) and 6 – (5 – x + 5 – y) = x + y – 4 units are to be transported from the factory at Q to the depots at B and C respectively. Thus, 5 – y ≥ 0 , x + y – 4 ≥0 i.e. y≤5 , x+y≥ 4 2019-20