Solving an Optimization Problem Using Implicit Differentiation

Solving an Optimization Problem using Implicit Differen­ tiation Suppose you wish to build a grain silo with volume V made up of a steel cylinder and a hemispherical roof. The steel sheets covering the surface of the silo are quite expensive, so you wish to minimize the surface area of your silo. What height and radius should the silo have for a given volume V ? Although it is possible to solve this problem by the same method used in the can design question, it turns out to be much simpler to use implicit differ­ d entiation to find dr SA. Answer this question for: a) a silo with a circular floor (to keep out gophers) and b) a silo with no built-in flooring (for use in regions with no gophers). Solution Our first task is to draw a sketch of the silo; this will help us in finding formulas for the total surface area and volume. The silo will have some radius r, and its cylindrical wall will have some height h. a) A silo including a circular floor: The surface area of the silo’s cylindrical wall is the circumference of the base times the height, or 2πrh; the surface area of the roof will be 12 (4πr2 ) and the floor will have surface area πr2 . Hence, the surface area of a silo with a floor is given by: SA = 2πrh + 2πr2 + πr2 = 2πrh + 3πr2 . 2 The volume of the�cylinder � will be πr h and the volume of the hemispherical 1 4 2 πr . The total volume of the silo will be: “attic” will be 2 3 2 V = πr2 h + πr3 . 3 V We could use this equation to determine that h = 2r 3 − πr 2 and then sub­ stitute this in to the expression for surface area, but the resulting equation d dr SA = 0 is unpleasant. Instead we apply implicit differentiation to the formula for surface area. SA = d SA dr = 2πrh + 3πr2 dh 2πh + 2πr + 6πr dr 1 We next use implicit differentiation on the formula for volume (or solve for h and differentiate normally) to find: V = dV dr = 0 −πr2 = dh dr dh dr = 2 πr2 h + πr3 3 dh 2 πr + 2πrh + 2πr2 dr dh 2 πr + 2πrh + 2πr2 dr 2πrh + 2πr2 = −2 · Plugging this in to the formula for d SA dr h+r . r d dr SA, we get: dh + 6πr dr � � h+r = 2πh + 2πr −2 · + 6πr r = 2πh − 4π (h + r) + 6πr = −2πh + 2πr. � d We see that dr SA = 0 exactly when h = r = 3 3V 5π . = 2πh + 2πr Is this a minimum? There are no other critical points, and the extreme outcomes are a tall, thin silo with an extremely large surface area� or a short, wide silo (as the height approaches 0 and the radius approaches 3 3V 2π ). After using the volume formula to find expressions for r in terms of V , we see that in the case h = 0, SA = = When h = r = � 3 3V 5π 2πrh + 3πr2 �� �2 3 3V 3π 2π � 2 3 243V π . 4 SA = SA = 2πrh + 3πr2 = 5πr2 �� , = = 5π √ 3 2 3 3V 5π 9V 2 π. �2 The solution in which r = h has the least surface area. Note that the silhouette of this silo is a rectangle with a semi-circle on top, which is closer to the shape of a water or oil tank than that of a silo. The tall, thin silos actually used to store grain take up less land area, leaving more room for farmland or other silos. b) A silo with an open floor: The formula for the volume of this silo is identical to the formula for the volume of one with floor included. 2 V = πr2 h + πr3 3 The surface area of the silo is given by adding the surface area of the wall, which is 2πrh, to the surface area of the roof, which is 2πr2 . SA = 2πrh + 2πr2 For practice, we’ll again use implicit differentiation to find The calculation of lation in part (a): dh dr SA = d SA dr = d dr SA: 2πrh + 2πr2 dh 2πh + 2πr + 4πr dr using implicit differentiation is identical to the calcu­ V = dV dr = 0 −πr2 dh dr dh dr = = 2 πr2 h + πr3 3 dh 2 πr + 2πrh + 2πr2 dr dh 2 πr + 2πrh + 2πr2 dr 2πrh + 2πr2 = −2 · h+r . r Plugging in, we find: d SA dr dh + 4πr dr � � h+r = 2πh + 2πr −2 · + 4πr r = 2πh − 4π (h + r) + 4πr = 2πh + 2πr = −2πh. 3 In this case the extreme solution in which h = 0 is actually the minimal surface area solution. We conclude that the minimum surface area silo con­ taining a given volume V is just a hemisphere, if the silo is built without a floor! 4