Question

Suppose that 320 feet of fencing are available to enclose a rectangular field and that one side of the field must be given double fencing. What are the dimenstions of the rectangle that will yield the largest possible area?

Answer 1

if l = length and w = width
then 3l + 2w = 320
2w = 320 - 3l
w = 160 - 1.5l
area A = w l = 160l - 1.5l^3
differentiate
dA/ dl = 160 - 3l
for max/min area da/dl = 0

160-3l = 0
l = 160/3
second derivative is -3 - negative so its a maxm

maximum area has length 53.33 and width = 160 - 1.5 * 160/3 = 80

dimensions are 80 * 53.33 feet

Answer 2

1) P = 2L + 3 W = 320
2) A = L * W

Solve for L or W

160 - 3/2 W = L
Substitute into equation 2
A(W) = 160W - 3/2 W^2

Maximum/Min will be found by the first derivative setting equal zero
A'(W) = 160 - 3W = 0
W = 160/3
L = 80 Dimensions would be 80, 160/3

Answer 3

pls reverse w and l

Answer 4

Thanks guys. Only question is how you came up with the 3w and 2l in the beginning?