[Max-min word problems examples] Example. A rectangular pen consists of two identical sections separated by a fence. 180 feet of fence is used for the outside of the pen and the dividing fence. What dimensions for each section produce the pen of largest area? Let each section be x feet in width and y feet in height. x x y y y x x The total area is A = 2xy. Since 180 feet of fence are available, 180 = 4x + 3y y= Plug this into A = 2xy: 1 (180 − 4x) 3 2 1 A = 2x · (180 − 4x) = (180x − 4x2 ). 3 3 The endpoints are given by the extreme cases x = 0 and y = 0; note that y = 0 gives 180 = 4x, Differentiate: A′ = Set A′ = 0 and solve for x: The maximum occurs when x = or x = 45. 2 (180 − 8x). 3 2 (180 − 8x) = 0 3 45 x= 2 x 0 A 0 45 2 1350 45 0 45 . In this case, 2 45 1 180 − 4 · = 30. y= 3 2 1 Example. The volume of a circular cylinder (with a top and a bottom) is 3456π. What values for the radius r and the height h give the smallest total surface area (the area of the side plus the area of the top plus the area of the bottom)? r h The area of the side is 2πrh, the area of the top is πr 2 , and the area of the bottom is πr 2 . The total surface area is A = 2πrh + 2πr 2 . The volume is 3456π = πr 2 h, Solving for h gives h = so 3456 = r 2 h. 3456 . Plug this into A: r2 A = 2πr · 6912π 3456 + 2πr 2 = + 2πr 2 . r2 r The only restriction on r is r > 0. Hence, r is not restricted to a closedinterval a ≤ r ≤ b. Therefore, I’ll use the Second Derivative Test. Compute A′ and A′′ : 6912π A′ = − 2 + 4πr, r A′′ = Set A′ = 0 and solve for r: − 13824π + 4π. r3 6912π + 4πr = 0 r2 4πr 3 = 6912π r 3 = 1728 r = 12 This gives h = Now 3456 = 24. 144 A′′ (k) = 13824π + 4π = 12π > 0. 1728 Hence, r = 12 is a local min. Since it’s the only critical point, it’s an absolute min. Example. A rectangular box with a square bottom and no top is made with 972 square inches of cardboard. What values of the length x of a side of the bottom and the height y give the box with the largest volume? 2 y x x The volume is V = x2 y. The area of the 4 sides is 4xy, and the area of the bottom is x2 . So 972 = 4xy + x2 . Solving for y gives y= 972 − x2 . 4x Plug this into V and simplify: V = x2 · 972 − x2 1 1 = x(972 − x2 ) = (972x − x3 ). 4x 4 4 Note that x 6= 0, since x = 0 plugged into 972 = 4xy + x2 gives 972 = 0 a contradiction. So the only restriction on x is that x > 0. Since x is not restricted to a closed interval [a, b], I’ll use the Second Derivative Test. Compute the derivatives: 1 V ′ = (972 − 3x2 ). 4 1 3 V ′′ = · (−6x) = − x. 4 2 Find the critical points: 1 (972 − 3x2 ) = 0 4 972 − 3x2 = 0 3x2 = 972 x2 = 324 x = ±18 Since x is a length, it must be positive, so x = 18. This gives y= 972 − 324 = 9. 72 Plug x = k into the Second Derivative: 3 V ′′ (k) = − · 18 = −27 < 0. 2 x = 18 is a local max, but it’s the only critical point, so it’s an absolute max. 3 Example. A rectangular sheet of cardboard is 16 wide and 21 inches high. Squares with sides of length x are cut out of each corner. The four resulting tabs are folded up to make a rectangular box (with no top). Find the value of x which gives the box of largest volume. x x x x 21 x x x x x 21-2x 16-2x 16 The base of the box is 16 − 2x by 21 − 2x, and the height is x. Hence, the volume is V = x(16 − 2x)(21 − 2x) = 4x3 − 74x2 + 336x. The endpoints are x = 0 (cut no squares out) and x = 8 (cut halfway from the bottom to the top). Obviously, these endpoints give volume 0, but it means that V is defined on a closed interval. So I just need to test the critical points on the interval and find the one that gives a max. Differentiate: V ′ = 12x2 − 148x + 336. Set V ′ = 0 and solve for the critical points. This gives √ 37 ± 361 x= . 6 28 The roots are x = and x = 3. 3 28 Now x = is greater than 8, so it’s outside the interval. So I only need to consider the other root 3 x = 3. x 0 x=3 8 V 0 450 0 Thus, x = 3 maximimizes the volume. Example. A rectangular poster is printed on a piece of cardboard with area 216 square inches. The printed region is a rectangular region centered on the cardboard. There are unprinted (blank) margins of width 3 inches on the left and right of the printed region, and width 2 inches on the top and bottom of the printed region. What dimensions for the printed region maximize its area? x+6 2 y 3 printed region x margin 2 4 3 y+4 Suppose the printed area is x by y. Its area is A = xy. Since there are margins of width 3 on the left and right, and margins of width 2 on top and bottom, the area of the cardboard is (x + 6)(y + 4) = 216. Solving for y yields y= 216 − 4. x+6 Plug this into the expression for A to obtain A=x 216x 216 −4 = − 4x. x+6 x+6 The extreme cases are x = 0 and y = 0. If y = 0, then (x + 6)(4) = 216 x = 48 So the endpoints are x = 0 and x = 48. (These make the area of the printed region 0, so they obviously don’t give the maximum!) Compute the derivative: A′ = 1296 (x + 6)(1) − (216x)(1) −4= − 4. (x + 6)2 (x + 6)2 Find the critical points: 1296 −4=0 (x + 6)2 1296 =4 (x + 6)2 1296 = 4(x + 6)2 324 = (x + 6)2 ±18 = x + 6 −6 ± 18 = x Since x is a length, it can’t be negative, so take the plus sign. This gives x = 18 − 6 = 12. Plug this into y = 216 − 4 to get y = 8. x+6 x 0 12 48 A 0 96 0 This shows that x = 12 and y = 8 maximize the area of the printed region. c 2014 by Bruce Ikenaga 5

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