OPTIMIZATION PROBLEM:

Question

"A rectangular field is to be enclosed on four sides with a fence. Fencing costs $7 per foot for two opposite sides, and $5 per foot for the other two sides. Find the dimensions of the field of area 620ft^2 that would be the cheapest to enclose".

Thank you in advance!

Answer 1

I forgot to include the answer choices:
A. 39 ft @ $7 by 17.8 ft @ $5
B. 21 ft @ $7 by 29.5 ft @ $5
C. 29.5 ft @ $7 by 21 ft @ $5
D. 17.8 ft @ $7 by 34.9 ft @ $5

Answer 2

If the dimensions are x and y, then
xy = 620
The perimeter is
p(x,y) = 2(x+y)
and the cost is thus c(x,y) = 2(7x+5y) = 14x + 10y
But, using y from above,
c(x) = 14x + 10(620/x) = 14x + 6200/x
Now minimize the cost by finding where dc/dx = 0

Answer 3

let each of the expensive sides by x ft, and each of the other two sides be y ft
xy = 620
y = 620/x

cost = 7(2x) + 5(2y)
= 14x + 10(620/x)
d(cost)/dx = 14 - 6200/x^2 = 0 for a min of cost
14 = 6200/x^2
x^2 = 3100/7
x = √(3100/7) = ...
carry on

Answer 4

c = 7(2x) + 5(2y) = 14 x + 10 y
620 = x y
so
y = 620/x

c = 14 x + 6200/x
dc/dx = 14 - 6200/x^2 which is zero for min c
14 x^2 = 6200
x^2 = 442,86
x = 21 at $7
then y = 620/21 = 29.5 at $5