Question

2. (10 points) There are $320 available to fence in a rectangular garden. The fencing for the side of the garden facing the road costs $6 per foot, and the fencing for the other three sides costs $2 per foot. Find the dimensions of the garden that maximize the area of the garden

Answer 1

(a) Let x be the length of the side facing the road. The cost of this side is $6 per foot so the cost for this side is 6.1 dollars. The other three sides are T. y, and y and have a cost of $2 per foot. The total cost of the other three sides is then 23+ y + y) = 2(x + 2y) = 2r + 4y dollars. The total cost of the fencing is then C = 6:1 + 2.1 + 4y = 8.1 + 4y dollars. We are given that the maximum must is $320 so the constraint equation is 320 = 8.1 +4y. The objective is to maximize the arca. The objective cquation is then A=.my. 2 (b) Solving the constraint equation 8x + 4y = 320 for y gives: 8x + 4y = 320 4y = 320 - 8.3 y = 80 - 2.2 Substitutung this into the objective equation then gives: A = XY = x(80 – 2x) = 80.0 - 2r2 3 (c) Find the derivative of A = 90.x – 2x2. Then set the derivative equal to O to find the critical value: A = 80 - x2r2 A' = 80 - 42 0 = 80 - 4.r 4.r = 80 X = 20 Since A' = -4 < 0 for all x, then r = 20 must be a maximum. The maximum area then occurs when r = 20 and y = 80 – 2.x = 80 – 2(20) 80 – 40 = 40.