diff cal

MATH 401 – DIFFERENTIAL CALCULUS Theorem4.3.1. If f(x) exists for all values of x in the open interval (a,b), and if f has a relative extremum at c, where a < c < b, and if f 'c exists, then f 'c  0 Illustration1: Let f be a function defined by f x  x2  4x  5 f 'x  2x  4 f '2  2(2)  4  4  4 f '2  0 Because f '2  0 , f may have a relative extremum at 2. Since f 2  1 and 1  f x where either x  2 or x  2 , definition4.6.2, guarantees that f has a relative minimum value at 2. Figure 5 shows the graph of f, a parabola whose vertex is at point (2,1) where the graph has a horizontal tangent. Note that f 'c  0 can equal zero even Figure 5 if f does not have a relative extremum at c, as shown in illustration 2. Illustration2: Let f be a function defined by f x  (x 1)3  2 Figure 6 f 'x  3(x 1)2 (1)  0 f 'x  3(x 1)2 f '1  3(11)2  3(0)2 f '1  0 Because f '1  0 , f may have a relative extremum at 1. However, because f 1  2 and 2  f (x) when x  1 and 2  f (x) when x  1, neither Definition 4.6.1 nor Definition 4.6.2 applies. So f does not have a relative extremum at 1. The graph of this function is shown in Figure 6 has a horizontal tangent at the point (1,2), which is consistent with the fact that the derivative is zero. 148

MATH 401 – DIFFERENTIAL CALCULUS Definition4.3.1.c Critical Number If c is a number in the domain of the function f , and if either f 'c  0 or f 'c does not exists, then c is a critical number of f. Example1. Find the critical numbers of f x  x4  4x3  2x2 12x Solution: We compute f 'x , set it equal to zero and solve for x. f x  x4  4x3  2x2 12x f 'x  4x3 12x2  4x 12 f 'x  0 4x3 12x2  4x 12  0 x3  3x2  x  3  0 (x3  3x2 )  (x  3)  0 x2 (x  3)  (x  3)  0 (x  3)(x2 1)  0 (x  3)(x 1)(x 1)  0 (x  3)  0 (x 1)  0 (x 1)  0 x  3 x 1 x  1 We have confirm that the critical numbers are -3, 1, and -1 Example2. Find the critical numbers of f x  x2ex Solution: We compute f 'x , set it equal to zero and solve for x. f x  x2ex f 'x  x2 (ex )  ex (2x) f 'x  x2ex  2xex f 'x  0 then  x2ex  2xex  0 ex (x2  2x)  0 ex  0 (x2  2x)  0 x(x  2)  0 (x  2)  0 x0 x2 We have confirm that the critical numbers are 0 and 2 149

MATH 401 – DIFFERENTIAL CALCULUS 4.3.1. Increasing and Decreasing Functions and the First-Derivative Test The term increasing, decreasing, and constant are used to describe the behavior of a function as we travel left to right along its graph. An example is shown below. Definition4.3.2.a Increasing Function A function f defined on an interval is increasing on that interval if and only if f x1   f x2  whenever x1  x2 , where x1 and x2 are any numbers in the interval. 150

MATH 401 – DIFFERENTIAL CALCULUS Definition4.3.2.b Decreasing Function A function f defined on an interval is decreasing on that interval if and only if f x1   f x2  whenever x1  x2 , where x1 and x2 are any numbers in the interval. Definition4.3.2.c Constant Function A function f defined on an interval is constant on that interval if and only if f x1   f x2  for all points x1 and x2 . 151

MATH 401 – DIFFERENTIAL CALCULUS Theorem4.3.2.a Let the function f be continuous on the closed interval [a,b] and differentiable on the open interval (a,b): (i) If f 'x  0 for every value of x in (a,b), then f is increasing on [a,b] (ii) If f 'x  0 for every value of x in (a,b), then f is decreasing on [a,b] (iii) If f 'x  0 for every value of x in (a,b), then f is constant on [a,b] Illustration1. Find the intervals on which f x  x2  4x  3 is increasing and decreasing. Solution: Graph the equation. Take the derivative of f x  x2  4x  3 f 'x  2x  4 thus  f 'x  0 when x  2  f is decreasing on - ,2   f 'x  0 when x  2  f is increasing on 2, Illustration2. Find the intervals on which f x  x3 is increasing and decreasing. Solution: Graph the equation. Take the derivative of f x  x3 f 'x  3x2 thus  f 'x  0 when x  0  f is increasing on - ,0   f ' x  0 when x  0  f is increasing on 0, 152

MATH 401 – DIFFERENTIAL CALCULUS Theorem4.3.2.b Suppose that f is a function defined on an open interval containing the point x0 . If f has a relative extremum at x  x0 , then x  x0 is a critical point of f ; that is, either f '(x0 )  0 or f is not differentiable at x0 . In general, we define a critical point for a function f to be a point in the domain of f at which either the graph of f has a horizontal tangent line or f is not differentiable (line is vertical). To distinguish between the two types of critical points we call x a stationary point of f if f '(x)  0 . Illustration1. Find all critical points of f (x)  x3  3x 1 Solution: We compute f 'x set it equal to zero and solve for x. Since f (x)  x3  3x 1 then f 'x  3x2  3 let f 'x  0 then 3x2  3  0 3(x2 1)  0  3(x 1)(x 1)  0 Thus, 3  0 , (x 1)  0  x  1 (x 1)  0  x 1 If x  1  y  3 and If x  1  y  1 Therefore the critical points are (1,3) and (1,1) Illustration2: Let f be a function defined by f x  x2  4x  5 Solution: We compute f 'x set it equal to zero and solve for x. Since f x  x2  4x  5 then f 'x  2x  4 let f 'x  0 then 2x  4  0 2(x  2)  0 Thus, 2  0 (x  2)  0 x2 If x  2  y 1 Therefore the critical point is (2,1) 153

MATH 401 – DIFFERENTIAL CALCULUS Theorem4.3.2.b asserts that the relative extrema must occur at critical points, but it does not say that a relative extremum occurs at every critical point. A function has a relative extremum at those critical points where f ' changes sign. Theorem4.3.2.c The First-Derivative Test Suppose that f is a continuous at a critical point x0 (i) If f '(x)  0 on an open interval extending left from x0 and f '(x)  0 on an open interval extending right from x0 , then f has a relative maximum at x0 . (ii) If f '(x)  0 on an open interval extending left from x0 and f '(x)  0 on an open interval extending right from x0 , then f has a relative minimum at x0 . (iii) If f '(x) has the same sign on an open interval extending left from x0 as it does on an open interval extending right from x0 , then f does not have a relative extremum at x0 . 154

MATH 401 – DIFFERENTIAL CALCULUS The above Theorem4.3.2.c simply say that for a continuous function, relative maxima occur at critical points where the derivative changes from (+) to (–) and relative minima where it changes from (–) to (+). Illustration1. Given f (x)  x3  6x2  9x  3 , find the relative maximum and relative minimum point and sketch the graph using the first derivative test. Solution: Plot the graph of the function and compute the derivative of f to determine the critical values. f (x)  x3  6x2  9x  3 f '(x)  3x2 12x  9 f 'x  3(x2  4x  3) f 'x  3(x 1)(x  3) The only critical numbers are those For which f '(x)  0 and f '(x) does not exists. 3(x 1)(x  3)  0 30 (x 1)  0  x  1 (x  3)  0  x  3 The critical numbers are 1 and 3. To determine whether f has a relative extremum at these numbers, we apply the first-derivative test and summarize the results in the table. Interval f (x)  x3  6x2  9x  3 f 'x  3(x 1)(x  3) Conclusion x  1 + f is increasing x 1 1 0 f has a relative maximum 1 x 3 - f is decreasing x3 -3 0 f has a relative minimum x3 + f is increasing The sign of f 'x changes sign from + to – at x  1  relative maximum 155

MATH 401 – DIFFERENTIAL CALCULUS The sign of f 'x changes sign from - to + at x  3  relative minimum 52 Illustration2. Given f (x)  3x 3 15x 3 , find the relative maximum and relative minimum point and sketch the graph using the first derivative test. Solution: Plot the graph of the function and compute the derivative of f to determine the critical values. 52 f (x)  3x 3 15x 3 f '(x)  3 5 x 5 1  15 2 x 2 1  3 3 3 3 2 1 f '(x)  5x 3 10x 3 1 f '(x)  5x 3 x  2 The only critical numbers are those For which f '(x)  0 and f '(x) does not exists. 1 5x 3 x  2  0 1 5x 3  0  x  0 and (x  2)  0  x  2 The critical numbers are 0 and 2. To determine whether f has a relative extremum at these numbers, we apply the first-derivative test and summarize the results in the table. Interval 52 1 Conclusion f (x)  3x 3 15x 3 f '(x)  5x 3 x  2 x0 + f is increasing x0 0 Does not exists f has a relative maximum 0 x2 - f is decreasing x2  93 4 0 f has a relative minimum x2 + f is increasing The sign of f 'x changes sign from + to – at x  0  relative maximum 156

MATH 401 – DIFFERENTIAL CALCULUS The sign of f 'x changes sign from - to + at x  2  relative minimum Illustration3. Given f (x)  x2ex , find the relative maximum and relative minimum point and sketch the graph using the first derivative test. Solution: Plot the graph of the function and compute the derivative of f to determine the critical values. f (x)  x2ex f '(x)  x2 (ex )  ex (2x) f '(x)  x2ex  2xex f '(x)  ex (x2  2x) The only critical numbers are those For which f '(x)  0 and f '(x) does not exists. ex (x2  2x)  0 ex  0 (x2  2x)  0  x(x  2)  0 x0 (x  2)  0  x  2 The critical numbers are 1 and 3. To determine whether f has a relative extremum at these numbers, we apply the first-derivative test and summarize the results in the table. Interval f (x)  x2ex f '(x)  ex (x2  2x) Conclusion x0 0 - f is decreasing x0 0 f has a relative minimum 4e 2 + f is increasing 0 x2 0 f has a relative maximum x2 - f is increasing x2 157

MATH 401 – DIFFERENTIAL CALCULUS The sign of f 'x changes sign from - to + at x  0  relative minimum The sign of f 'x changes sign from + to - at x  2  relative maximum Exercise 3. Solve the following: (a) find the relative extrema of the function using the first-derivative test; (b) determine the values of x at which the relative extrema occur; (c) determine the intervals on which the function is increasing and decreasing; (d) Sketch the graph of the function from your answers. 1. f (x)  x3  3x2  3x 2. y  x3  3x2  5 3. f (x)  x2  4x  5 x2 4. f (x)  2x2 ln x 5. y  ex sin x 158

MATH 401 – DIFFERENTIAL CALCULUS 4.3.2. Concavity, Points of Inflection, and the Second-Derivative Test Although the sign of the derivative of f reveals where the graph of f is increasing or decreasing, it does not reveal the direction of the curvature. Figure4.3.3 below suggests two ways to characterize the concavity of a differentiable f on an open interval:  f is concave up on an open interval if its tangent lines have increasing slopes on that interval and is concave down if they have decreasing slopes.  f is concave up on an open interval if its graph lies above its tangent lines and concave down if it lies below its tangent lines. Figure4.3.3 Definition4.3.3.a Concave Upward and Downward If f is differentiable on an open interval, then f is said to be concave upward on the open interval if f ' is increasing on that interval, and f is said to be concave downward on the open interval if f ' is decreasing on that interval. Since the slopes of the tangent lines to the graph of a differentiable function f are the values of its derivative f ' , it follows from Theorem4.3.2.a (applied to f ' rather than f ) that f ' will be increasing on intervals where f '' is positive and that f ' will be decreasing on intervals where f '' is negative. Thus we have the following theorem 159

MATH 401 – DIFFERENTIAL CALCULUS Theorem4.3.3.a Let f be twice differentiable on an open interval. (i) If f ''  0 for every value of x in the open interval, then f is concave upward on that interval. (ii) If f '' 0 for every value of x in the open interval, then f is concave downward on that interval. Illustration1. Let f x  x2  4x  5 , determine where the graph is concave upward and concave downward. Solution: Given f x  x2  4x  5 f 'x  2x  4 f ''x  2 f 'x  0 then 2x  4  0 , solve for critical numbers 2(x  2)  0 Thus, 2  0 and (x  2)  0  x  2 The critical number is 2, summarized the results in Table below and graph Interval f (x) f 'x Conclusions f ''x Conclusions x2 - f is decreasing + f is concave upward x2 1 0 f has a relative + f is concave upward minimum x2 + f is increasing + f is concave upward 160

MATH 401 – DIFFERENTIAL CALCULUS Illustration2. Let f (x)  x3  3x 1 , determine where the graph is concave upward and concave downward. Solution: Given f (x)  x3  3x 1 f 'x  3x2  3 and f ''x  6x f 'x  0 then 3x2  3  0 , solve for critical numbers 3(x2 1)  0  3(x 1)(x 1)  0 Thus, 3  0 , (x 1)  0  x  1 (x 1)  0  x 1 The critical numbers are -1 and 1, summarized the results in Table below and graph. Interval f (x) f 'x Conclusions f ''x Conclusions x  1 f is increasing + f has a relative - f is concave downward x  1 30 maximum - f is concave downward 1 x 1 - f is decreasing - f is concave downward + then concave upward f has a relative x  1 -1 0 minimum + f is concave upward x 1 + f is increasing + f is concave upward 161

MATH 401 – DIFFERENTIAL CALCULUS Points where the curve changes from concave upward to concave downward or vice-versa are called points of inflection. Definition4.3.3.b Points of Inflection If f is continuous on an open interval containing a value x0 , and if f changes the direction of its concavity at the point x0 , f x0  , then we say that f has an inflection point at x0 , and we call the point x0 , f x0  on the graph of f an inflection point of f (see figure below) Figure4.3.3.b Example1. Given the equation f x  x3  3x2 1 determine the intervals on which f is increasing, decreasing, concave upward and concave downward. Locate all inflection points and confirm that your conclusions are consistent with the graph. Solution: Given f x  x3  3x2 1 f 'x  3x2  6x f ''x  6x  6 f 'x  0 then 3x2  6x  0 , solve for critical numbers 3x(x  2)  0 Thus, 3x  0  x  0 (x  2)  0  x  2 The critical numbers are 0 and 2, summarized the results in the Table and graph. 162

MATH 401 – DIFFERENTIAL CALCULUS Interval f (x) f 'x Table1 x0 1 + Conclusions x0 -3 0 0 x2 - f is increasing x2 0 f has a relative maximum x2 + f is decreasing f has a relative minimum f is increasing Interval f (x) f ''x Table2 x0 1 - Conclusions x0 -1 - 0 x 1 -3 - f is concave downward x 1 0 f is concave downward 1 x  2 + f is concave downward x2 + f has a point of inflection x2 + f is concave upward f is concave upward f is concave upward Table2 shows that there is a point of inflection at x  1, since f changes from concave downward to concave upward at that point. The point of inflection is (1,-1). 163

MATH 401 – DIFFERENTIAL CALCULUS Example2. Given the equation f x  x4  4x3 determine the intervals on which f is increasing, decreasing, concave upward and concave downward. Locate all inflection points and confirm that your conclusions are consistent with the graph. Solution: Given f x  x4  4x3 f 'x  4x3 12x2 and f ''x 12x2  24x f 'x  0 then 4x3 12x2  0 , solve for critical numbers 4x2 (x  3)  0 Thus, 4x2  0  x  0 and (x  3)  0  x  3 Thus critical numbers are 0 and 3 Table2 Interval f (x) f 'x Conclusions f ''x Conclusions f is concave upward x0 - f is decreasing + f has a point of inflection x0 0 0 f has neither a relative 0 -16 max nor min f is concave downward 0 x2 -27 - f has a point of inflection x2 - f is decreasing 0 + f is concave upward 2 x3 - f is decreasing + f is concave upward x3 - f is decreasing + f is concave upward x3 0 f has a relative minimum + f is increasing Table2 shows that there is a point of inflection at x  0 and x  2 since f changes from concave upward to concave downward and vice versa at that point. The point of inflection are (0,0) and (2,-16). 164

MATH 401 – DIFFERENTIAL CALCULUS Theorem4.3.3.b Suppose the function f is differentiable on some open interval containing c, and c, f c is a point of inflection of the graph of f . Then if f ''c exists, f ''c  0 Illustration1. Given the equation f x  x4 determine the intervals on which f is increasing, decreasing, concave upward and concave downward. Locate all inflection points if any and confirm that your conclusions are consistent with the graph. Solution: Given f x  x4 f 'x  4x3 and f ''x 12x2 f 'x  0 then 4x3  0  4x3  0  x  0 The critical number is 0, summarized the results in the Table and graph. Table1 Interval f (x) f 'x Conclusions f ''x Conclusions x0 x0 - f is decreasing + f is concave upward x0 0 0 f has a relative 0 f has no inflection point minimum + f is increasing + f is concave upward Table1 shows that there is no change in concavity and hence no inflection point at x  0 , even though f ''x  0 165

MATH 401 – DIFFERENTIAL CALCULUS Illustration2. Let f (x)  xex , determine the intervals on which f is increasing, decreasing, concave upward and concave downward. Locate all inflection points if any and confirm that your conclusions are consistent with the graph. Solution: Given f (x)  xex f 'x  xex  ex f 'x  x(ex )  ex (1) f ''x  (x)(ex )  ex (1)  (ex ) f 'x  xex  ex f ''x  xex  ex  ex f 'x  0 then  xex  ex  0 , f ''x  xex  2ex f ''x  0 solve for critical numbers  xex  ex  0  ex (x 1)  0 xex  2ex  0  ex (x  2)  0 Thus, ex  0 , (x 1)  0  x  1 Thus, ex  0 , (x  2)  0  x  2 (1,e1) critical point (2,2e2 ) possible inflection point Interval f (x) f 'x Conclusions f ''x Conclusions x 1 + f is increasing - f is concave downward f has a relative x  1 e1 0 - f is concave downward maximum 1 x  2 2e 2 - f is decreasing - f is concave downward x2 - f is decreasing 0 f has a point of inflection x2 - f is decreasing + f is concave upward 166

MATH 401 – DIFFERENTIAL CALCULUS There is another test for relative extrema that is based on the following geometric observation: • A function f has a relative maximum at stationary point if the graph of f is concave downward on an open interval containing that point. • A function f has a relative minimum at stationary point if the graph of f is concave upward on an open interval containing that point. Theorem4.3.3.b Second Derivative Test Suppose that f is twice differentiable at the point x0 (i) If f 'x0   0 and f ''x0   0 , then f has a relative minimum at x0 . (ii) If f 'x0   0 and f ''x0   0 , then f has a relative maximum at x0 . (iii) If f 'x0   0 and f ''x0   0 , then the test is inconclusive; that is, f may have a relative maximum, a relative minimum, or neither at x0 . Illustration1. Given f x  3x5  5x3 Find the relative extrema of f by applying the second-derivative test. Use this information to sketch the graph of f. Solution: Given f x  3x5  5x3 f 'x  15x4 15x2 and f ''x  60x3  30x f 'x  0 then 15x4 15x2  0  15x2 (x2 1)  0  15x2 (x 1)(x 1)  0 15x2  0 , (x 1)  0 , (x 1)  0 x  0 , x  1, x 1 167

MATH 401 – DIFFERENTIAL CALCULUS Table1 Interval f (x) f 'x f ''x Conclusions x  1 x0 20 - f has a relative maximum x 1 00 0 f has neither a relative maximum nor minimum -2 0 + f has a relative minimum Illustration2. Given 3y  x3  3x2  9x  3 Find the relative extrema of f by applying the second-derivative test. Use this information to sketch the graph of f. Solution: Given 3y  x3  3x2  9x  3 y  1 x3  x2  3x 1 3 y  f x f x  1 x3  x2  3x 1 3 f 'x  x2  2x  3 and f ''x  2x  2 f 'x  0 then x2  2x  3  0  (x  3)(x 1)  0 (x  3)  0 , (x 1)  0 x  3 , x 1, 168

MATH 401 – DIFFERENTIAL CALCULUS Table1 Interval f (x) f 'x f ''x Conclusions x  3 10 0 - f has a relative maximum x 1 2 0 + f has a relative minimum 3 4.3.3. Sketching Graphs (Curve Tracing) of Algebraic and Transcendental Functions We learned in the previous section how properties of graphs of functions can be determined from their derivative. Now, we summarized the steps incorporating the properties discussed in this chapter that we should follow when sketching the graph of a function f. 1. Find any x and y intercepts. 2. Compute for the first and second derivative. 3. Determine the critical numbers of f. These are the values of x in the domain of f for which either f 'x  0 or f 'x does not exists. 4. Determine the relative extremum using the first-derivative test or second-derivative test. 5. Determine the intervals on which f is increasing or decreasing. 6. Find critical numbers of f ' , that is, the values of x for which f ''x does not exists or f ''x  0 to obtain possible inflection points. 169

MATH 401 – DIFFERENTIAL CALCULUS 7. Check for concavity of the graph. Example1. Analyze and trace the curve of y  4x 4  x2 Solution: The x and y intercept at the origin (0,0) Since y f x then f x  4 4x  x2 f 'x  (4  x2 )(4)  (4x)(2x)  16  4x2  8x2  16  4x2 (4  x2 )2 (4  x2 )2 (4  x2 )2 f 'x  16  4x2 let f 'x  0 (4  x2 )2 16  4x2 0  16  4x2 0  4(4  x2 )  0  4(2  x)(2  x)  0 then (4  x2 )2 4  0, (2  x)  0  x  2 , (2  x)  0  x  2 Critical values are -2 and 2. Critical points are (-2,-1) and (2,1) ''x  (4  x2 )2 (8x)  (16  4x2 )(2)(4  x2 )(2x) (4  x2)2 2  f f ''x  (8x)(4  x2 )2  4x(16  4x2 )(4  x2 ) (4  x2 )4   f ''x  (4x)(4  x2 ) (2)(4  x2 )  (16  4x2 ) (4  x2 )4  f ''x  (4x) 8  2x2  16  4x2 ) (4  x 2 )3 f ''x  (4x)(2x2  24)  (4x)(2)(x2 12) (4  x2 )3 (4  x2 )3 f ''x  (8x)(x2 12) let f ''x  0 (4  x2 )3 (8x)(x2 12)  0  (8x)(x2 12)  0 then (4  x2 )3 8x  0  x  0 (x2 12)  0  x2  12  x  2 3 Possible point of inflections are 0,0,   2 3,  2 3  and  2 3, 3  2 170

MATH 401 – DIFFERENTIAL CALCULUS Interval f (x) f 'x Conclusions f ''x Conclusions f is decreasing x  2 3 3 - - f is concave downward x  2 3 2 f is decreasing  2 3  x  2 - 0 f has inflection point f is decreasing x  2 -1 - f has a relative + f is concave upward 0 + f is concave upward 2 x0 0 + minimum + f is concave upward x0 + f is increasing 0 f has inflection point + f is increasing - f is concave downward 0 x2 0 f is increasing - f is concave downward - f has a relative - f is concave downward x2 1 - maximum 0 f has inflection point 2x2 3 3 f is decreasing x2 3 2 - + f is concave upward x2 3 f is decreasing f is decreasing Example2. Analyze and trace the curve of f x  3x5  5x4 Solution: Given f x  3x5  5x4 f 'x  15x4  20x3 let f 'x  0 15x4  20x3  0  5x3 (3x  4)  0 then 5x3  0  x  0 , (3x  4)  0  x   4 3 171

MATH 401 – DIFFERENTIAL CALCULUS Critical values are 0 and  4 . Critical points are (0,0) and   4 , 256  3  3 81  f ' ' x  60x3  60x2 let f ''x  0 60x3  60x2  0  (60x2 )(x 1)  0 then 60x2  0  x  0 (x 1)  0  x  1 Possible point of inflections are 0,0, and 1,2 Interval f (x) f 'x Conclusions f ''x Conclusions x 4 256 + f is increasing - f is concave downward 3 81 x4 3 0 f has a relative maximum - f is concave downward 2  4 3  x  1 0 - f is decreasing - f is concave downward x  1 - f is decreasing 0 f has inflection point - f is decreasing + f is concave upward 1 x  0 0 f has a relative minimum 0 x0 + f is increasing + f is concave upward x0 172

MATH 401 – DIFFERENTIAL CALCULUS Exercise 4. Solve the following: (a) find the relative extrema of the function using the second-derivative test; (b) determine the values of x at which the relative extrema occur; (c) determine the intervals on which the function is increasing and decreasing; (d) concave upward or concave downward; (e) Locate inflection points if any and (f) Sketch the graph of the function from your answers. 1. f (x)  2x3  3x2 12x  7 2. y 2 x2  4 3. y  3x4  4x3 12x2 4. y  x 1 x2 5. y  1 x5  2 x3 53 173

MATH 401 – DIFFERENTIAL CALCULUS 4.4. OPTIMIZATION PROBLEMS One of the most common applications of calculus involves the determination of minimum and maximum values. As a would-be engineer, you will encounter terms like greatest strength, greatest voltage, greatest profit, greatest distance, optimum size, least size, least time, and least cost. Finding solution to these situations are called Optimization Problems. To solve this, we need to be familiar on the application of calculus in solving these optimization problems. Here are the guidelines for solving maximum and minimum problems. GUIDELINES FOR SOLVING OPTIMIZATION PROBLEMS 1. Identify all given quantities and all quantities to be determined. If possible, make a sketch. 2. Write a primary equation for the quantity that is to be maximized or minimized. (Review useful formulas from Geometry) 3. Reduce the primary equation to one having a single independent variable. This may involve the use of secondary equations relating the independent variables of the primary equation. 4. Determine the feasible domain of the primary equation. That is, determine the values for which the stated problem makes sense. 5. Determine the desired maximum or minimum value by the calculus techniques discussed in the previous Section 4.2. TAKE NOTE: When performing Step 5, recall that to determine the maximum or minimum value of a continuous function f on a closed interval, you should compare the values of f at its critical numbers with the values of f at the endpoints of the interval. Another way is to apply the Second Derivative Test. 4.4.1 Optimization Problems involving Algebraic Functions The following examples illustrate the applications of calculus in solving optimization problems involving algebraic functions: Example 4.4.1.1 Maximum Volume of a Box An engineering firm wants to design an open box having a square base and a surface area of 192 square inches, as shown in the figure. What dimensions will produce a box with maximum volume? 174

MATH 401 – DIFFERENTIAL CALCULUS Solution: Step 1: Given a box with a square base, let x be its length and width and y be its height. Given its surface area . Surface Area = 192 Figure 1. Open box with a square base Step 2: Since the box has a square base, its Volume is This is a primary equation because it gives a formula for the quantity to be maximized. Step 3: Write V as a function of just one variable, say x. To do this we need a secondary equation which is the surface area of the box. Solve this equation for h in terms of x to obtain . Substituting into the primary equation for V produces Step 4: Determine the feasible domain of x that will yield a maximum value of V. That is, what values of x make sense in this problem? We all know that the volume , and x must be nonnegative and that the area of the base of the box is at most 192. So, the feasible domain is . Step 5: To maximize V, find the critical numbers of the volume function on the interval . To find the critical values of x, set 175

MATH 401 – DIFFERENTIAL CALCULUS These are the critical numbers. We will not consider the because it is not part of the feasible domain. To verify if this will give a maximum volume, apply the Second Derivative Test. Substituting gives Since , the V has a relative maximum. Therefore we conclude that the dimensions of the box with maximum volume is when its length or width . When , the height of the box . The maximum volume of . the box is Example 4.4.1.2 Minimum Material for a Piece of Paper A rectangular page is to contain 24 square inches of print. The margins at the top and bottom of the page are to be 1.5 inches, and the margins on the left and right are to be 1 inch. What should the dimensions of the page be so that the least amount of paper is used? Solution: Step 1: Start by labeling a drawing with the known and unknown quantities. 11.5 . 5 1.5 Figure 2. Rectangular Page 176

MATH 401 – DIFFERENTIAL CALCULUS Step 2: Let A be the area which is the quantity to be minimized. The area of the whole page which is the primary equation is given as Step 3: Write A as a function of one variable only, let us say x. To do this we need a secondary equation which is the area of print; 24 square inches. This is given by: Solving this equation for y gives . Now, substituting it into the primary equation produces Determine the feasible domain of x that will yield a maximum value of A. We all know that the area A , and x must be nonnegative, that is Step 5: To minimize A, find the critical values of x by differentiating A with respect to x Set These are the critical numbers. We will not consider the because it is not part of the feasible domain. To verify if this will give a minimum area, apply the Second Derivative Test. 177

MATH 401 – DIFFERENTIAL CALCULUS Substituting gives Since , then A has a relative minimum when Therefore, we concluded that the dimensions of the page with minimum area should be by Example 4.4.1.3 Endpoint Maximum Four feet of wire is to be used to form a square and a circle. How much of the wire should be used for a square and how much should be used for the circle to enclosed the maximum total area? Solution: Step 1: Given the total length of the wire is 4 ft. This is equal to the sum of perimeter of a square and the circumference of a circle. x Perimeter = 4x x Area = x2 4 feet .r Circumference = 2πr Area = πr2 Figure 3. A square and a circle Step 2: Let A be the total area which is the quantity to be maximized. The rimary equation is given as Step 3: Write A as a function of one variable only, let us say x. To do this we need a secondary equation which is the total length of the wire = 4ft. This is given by: Solving for r, Now, substituting it into the primary equation produces 178

MATH 401 – DIFFERENTIAL CALCULUS Determine the feasible domain of x that will yield a maximum value of A. The area A , and restricted by the square’s perimeter. Step 5: To maximize A, find the critical values of x by differentiating A with respect to x Set This is the only critical number in the feasible domain. To verify if this will give a maximum area, apply the Second Derivative Test. Since , then A has a relative minimum when But what we need is the value x that will give a maximum area. So, using the endpoints in the feasible domain and the critical value, the respective total area are given as: Therefore we concluded that the value of x that will give maximum area is when That is, all the wire is used for the circle. 179

MATH 401 – DIFFERENTIAL CALCULUS Example 4.4.1.4 Maximum Illumination A Norman window consists of a rectangle surmounted by a semicircle. If the perimeter of a Norman window is to be 40 ft, determine what should be the radius of the semicircle and the height of the rectangle such that the window will admit the most light? Solution: Step 1: Draw a picture and label the variables: x = length of a rectangle x = diameter of a semicircle x/2 = radius of a semicircle y = height of a rectangle Figure 4. The Norman window Step 2: The primary equation for this problem is area - the area for this window is the rectangular section plus the semicircular section. It is given by: Step 3: Write A as a function of one variable only, let us say x. To do this we need a secondary equation which is the perimeter - three sides of the rectangle plus the semicircle. It is given by: Solving for y in terms of x, it will probably make things easier if we multiplied both sides by 2 first to get rid of that fraction on the right side: 180

MATH 401 – DIFFERENTIAL CALCULUS Substituting this into the primary equation gives Determine the feasible domain of x that will yield a maximum value of A. The area A , and x must be nonnegative. Step 5: To maximize A, find the critical values of x by differentiating A with respect to x Set This is the only critical number in the feasible domain. To verify if this will give a maximum area, apply the Second Derivative Test. 181

MATH 401 – DIFFERENTIAL CALCULUS Since , then A has a relative maximum. We can now concluded that the radius of the semicircle is , and the height of the rectangle is Example 4.4.1.5 Minimum Surface Area A 120-m3 closed aluminum tank is to be in the form of a right-circular cylinder, determine the base radius of the tank if the least amount of aluminum is to be used in its manufacture. Solution: Step 1: Draw the figure of a right-circular cylindrical tank. We are asked to determine the base radius for which the total surface area of the tank is a minimum. r = base radius h = height V = Volume = 120 m3 Figure 5. Cylindrical Tank Step 2: The primary equation for this problem is total surface area which is equal to the lateral surface area and the total area of both the top and bottom (each of . If S is the total surface area, then Step 3: Write S as a function of one variable only, let us say r. To do this we need a secondary equation which is the volume of a right-circular cylinder. It is given by: V 182

MATH 401 – DIFFERENTIAL CALCULUS Solving this equation for h, we have , and substituting into the primary equation, we obtain S as a function of r: Determine the feasible domain that will yield a minimum value of S. From the equation defining S, r cannot be 0. Theoretically, r may be any positive number. Therefore, the feasible domain of S is . Step 5: To minimize S, find the critical values of r by differentiating S with respect to r, we have Set This is a critical number in the feasible domain. To verify if this will give a minimum value of S, apply the Second Derivative Test. If we substitute the critical value of r, then we will get . Therefore, S has a relative minimum value. We can now conclude that the least amount of Aluminum will be used in the manufacture of the tank when the base radius is 183

MATH 401 – DIFFERENTIAL CALCULUS Example 4.4.1.6 Minimum Cost of Box Material A closed box with a square base is to have a volume of 4,000 cubic centimeters. The material for the top and the bottom of the box is to cost 6 cents per square centimeter and the material for the sides is to cost 3 cents per square centimeter. Find the dimensions of the box so that the total cost of the material is least. Solution: Step 1: Draw the figure and label the quantities: x = length of a side of the square base y = depth of the box y x V = Volume = 4,000 cm3 Figure 6. A closed Box Step 2: The primary x for this problem is total cost of the material plus 3 cents equation which is 6 cents x combined area of the top and bottom x area of sides(4xy . The number of cents in the total cost of the material is Step 3: Write C as a function of one variable only, let us say x. To do this we need a secondary equation which is the volume of box that is equal to area of the base and the depth. V Solving this equation for y, we have , and substituting into the primary equation, we obtain C as a function of x: Determine the feasible domain that will yield a minimum value of C. From the equation defining C, x cannot be 0. Theoretically, x may be any positive number. Therefore, the feasible domain of C is . Step 5: To minimize C, find the critical values of x by differentiating C with respect to x, we have 184

MATH 401 – DIFFERENTIAL CALCULUS Set This is a critical number in the feasible domain. To verify if this will give a minimum value of C, apply the Second Derivative Test. If we substitute the critical value of , then we will get . Therefore, C has a relative minimum value. We can now conclude that the cost of the material will be least when side of a square base is and the depth is Example 4.4.1.7 Greatest Volume Find the dimensions of the right-circular cylinder of greatest volume that can be inscribed in a right-circular cone with a radius of 15cm and a height of 36cm. Solution: Step 1: Draw the figure and label the quantities: For the cylinder: x = radius h-y y = height For the cone: F r = base radius = 15 cm h = height = 36 cm Figure 7. Cylinder inscribed in a cone 185

MATH 401 – DIFFERENTIAL CALCULUS Step 2: The primary equation for this problem is volume of the cylinder which is Step 3: Write V as a function of one variable only, let us say x. To do this we need a secondary equation involving x and y. From Figure 7, and by similar triangles, and , Solving this equation for y, we have , and substituting into the primary equation, we obtain V as a function of x: Determine the feasible domain that will yield a maximum value of V. From the equation defining C, the feasible domain of C is . Step 5: To maximize V, find the critical values of x by differentiating V with respect to x, we have Set Solve for x These are the critical values of x, both of which are in . To verify which of these values will give a maximum value of V, apply the Second Derivative Test. Substitute the critical values: and 186

MATH 401 – DIFFERENTIAL CALCULUS When , . Therefore, V has a relative maximum value. Also the corresponding height is We can now conclude that the greatest volume of an inscribed cylinder in a given cone is , which occurs when the radius is 10 cm and the height is 12 cm. Example 4.4.1.8 is closest to the point MINIMUM Distance Which point on the graph of Solution: Step1: Sketch a graph of the function and label the points: Figure 8 Step 2: Since the function is symmetrical, and the point (0, 1) is in the middle, there are two points that have the same minimum distance. For this problem, we are minimizing distance, so for our primary equation we will use the formula for distance between two points, those points being (0,1) and (x, y): Step 3: Write d in terms of one variable only, say x. We need here the secondary equation which is the original function, It is already solved for y, so we can replace the y in the primary equation with it: 187

MATH 401 – DIFFERENTIAL CALCULUS Step 4: Determine the feasible domain that will yield a minimum value of d. Because d is smallest when the expression inside the radical is smallest, we need only to find the critical numbers of . The domain of this f(x) is the entire real line. Step 5: To minimize d, find the critical values of the function by differentiating f(x) with respect to x: Set Solving for x values, we have and , then and Substituting this to the original function, the corresponding y values are: From looking at the graph, we can see that x = 0 yields a relative maximum, which is not what is required in the problem. To verify if the critical values yield a minimum distance d, apply the Second Derivative Test 188

MATH 401 – DIFFERENTIAL CALCULUS Substituting the critical values, we have When , Since , then the d has a relative maximum. This means that the point is the farthest point from (0,1) which is not what is required in the problem. When , When , Since , then the d has a relative minimum. So, the closest points from point (0,1) are and . 4.4.2 Optimization Problems involving Transcendental Functions Example 4.4.2.1 Largest Lateral Surface Area A right-circular cylinder is to be inscribed in sphere of given radius. Find the ratio of the height to the base radius of the cylinder having the largest lateral surface area. Solution: Step 1: Draw the figure and label all the quantities: r = constant radius of sphere R = radius of cylinder h = height of cylinder θ θ = angle at the center of the sphere subtended by R Figure 9. Cylinder inscribed in a sphere 189

MATH 401 – DIFFERENTIAL CALCULUS Step 2: The primary equation for this problem is lateral surface area (S) of the cylinder, W v θT w need secondary equations expressing R and h in terms of θ F figure, consider a right triangle ABO, applying SOHCAHTOA, we have Substituting these into the primary equation, we obtain S as a function of θ: Determine the feasible domain that will yield a maximum value of S. From the equation defining S, the feasible domain is . Tz v θ differentiating S with θ take note here that r is a constant radius of a sphere, we have Set This is a critical number in the feasible domain 190

MATH 401 – DIFFERENTIAL CALCULUS To verify if this will give a maximum value of S, apply the Second Derivative Test. If we substitute the critical value of , then we will get a relative maximum value. . Since then S has When , the height h and the base radius R is We can now conclude that for the cylinder having the largest lateral surface area, the ratio of the height to the base radius is 2. Example 4.4.2.2 Best View An engineer designed a sports arena where the television screen is vertical and .4 m high. The lower edge is 8. m above an observer’s eye level. If the best view of the screen is obtained when the angle subtended by the screen at eye level is a maximum, how far from directly below the screen must the observer be? Solution: Step 1: Draw the figure and label all the quantities. angle subtended by the screen at eye level x = distance between the point directly below the screen and the observer Figure 10. Television screen 191

MATH 401 – DIFFERENTIAL CALCULUS Step 2: The primary equation for this problem is the angle subtended by the screen at eye level. From the figure, it is Step 3: Write as a function of one variable only, let us say x. To do this we need secondary equations that will relate and with x. Consider two right triangles in the figure and apply SOHCAHTOA. This gives Substituting these into the primary equation, we obtain as a function of x: Determine the feasible domain that will yield a minimum value of . From the equation defining , x cannot be 0. Theoretically, x may be any positive number. Therefore, the feasible domain is . Step 5: To maximize , find the critical values of x by differentiating with respect to x, we have Combining the right side by getting the LCD, we have 192

MATH 401 – DIFFERENTIAL CALCULUS Set These are the critical numbers. We will disregard the because it is not in the feasible domain. To verify if this will give a maximum value of , apply the First Derivative Test. For , the , so is increasing For , the , so has a relative maximum value For , the , so is decreasing We can now conclude that to have the best view of the screen, the observer should stand approximately 9.63 m from the point directly below the screen. Example 4.4.2.3 Largest Rectangle A computer is programmed to inscribe a series of rectangles in the first quadrant under the curve of . What is area of the largest rectangle that can be inscribed? Solution: Step 1: Sketch the graph and label the quantities. 193

MATH 401 – DIFFERENTIAL CALCULUS Figure 11. One of the possible rectangles under the curve. Step 2: The primary equation for this problem is area - the area of the rectangle which is given by: Step 3: Write A as a function of one variable only, let us say x. To do this we need a secondary equation which is the exponential curve . Substituting this into the primary equation gives Determine the feasible domain of x that will yield a maximum value of A. The area A , and x must be nonnegative. Step 5: To maximize A, find the critical values of x by differentiating A with respect to x. Use Product Rule Set This expression only equals zero when . 194

MATH 401 – DIFFERENTIAL CALCULUS To verify if this will give a maximum area, apply the Second Derivative Test. Use Product Rule to get the 2nd derivative If substitute the critical number , we get . Since , then A has a relative maximum. The length of the largest rectangle is , then its height is . We can now conclude that the area of the largest rectangle that can be inscribed is 4.4.3. Number Problems In this section, we are to apply optimization in finding two positive numbers that satisfy the given requirements. Example 4.4.3.1 The product is 100 and the sum is a minimum. Solution: Let x be the first number and y be the second number. Their product is while their sum Express S in terms of one variable only, say x. To do this, solve for y in the primary equation . Substitute this in their sum S, we have now S as a function of x: To minimize S, find the critical numbers by differentiating S with respect to x. 195

MATH 401 – DIFFERENTIAL CALCULUS Set S’ = 0 Since the requirement are two positive numbers, the numbers that will give a product of 100 and the sum is minimum are both 10. Example 4.4.3.2 The sum is 12 and their product is a maximum. Ans. 6 Solution: Let x be the first number and y be the second number. Their sum is while their product is Express P in terms of one variable only, say x. To do this, solve for y in the primary equation . Substitute this in their product P, we have now P as a function of x: To maximize P, find the critical numbers by differentiating P with respect to x. Set P’ = 0 then Since the requirement are two positive numbers, the numbers that will give a sum of 12 and the product is maximum are both 6. 196

MATH 401 – DIFFERENTIAL CALCULUS Exercises 4.4 Solve the following optimization problems: 1. A funnel of specific volume is to be in a shape of a right-circular cone. Find the ratio of the height to the base radius if the least amount of material is to be used in its manufacture. Ans. 2. A right-circular cone is to be circumscribed about a sphere of given radius. Find the ratio of the altitude to the base radius of the cone of least possible volume. Ans. 3. Find the volume of the largest right-circular cylinder that can be inscribed in a right-circular cone having a radius of 4 cm and a height of 8 cm. Ans. 4. A piece of wire 10 ft long is cut into two piece. One piece is bent into the shape of a circle and the other into a shape of a square. How should a wire be cut so that (a) the combined area of the two figures is as small as possible; (b) the combined area of the two figures is as large as possible? Ans. (a) radius of circle is (b) radius of circle is length of side of square is there is no square 5. A piece of wire 20 cm long is cut into two pieces, and each piece is bent into a shape of a square. How should a wire be cut so that the total area of the two squares is as small as possible? Ans. Cut in half 6. If R meters is the range of a projectile, then , , where ft/s is the initial velocity, m/s2 is the acceleration due to gravity, and is the radian measure of the angle that the launcher makes with the horizontal. Find the value of that makes the range maximum. Ans. 7. Which points on the graph of are closest to the point (0,2)? Ans. and 197