A farmer wants to keep his animals out of the water

Question

A farmer was told by the municipality to enclose a field by a river to keep his animals out of the water. He decides to enclose a rectangular field as show below. The fence on the river side cost $20 per foot all other side cost $4 dollars per foot

If he must enclose 97200 square feet, what values of x and y will minimize the cost?

Answer 1

I presume that the y dimension has the river on one side.

Area = xy = 97200

y = 97200/x

Cost = 20y + 4y +4*2*x =24y + 8x

Note: only one side is bounded by the river.

Cost = 24*97200/x + 8x

Take the derivative of both sides:

0 = -24*97200/x² + 8

Multiplying both sides by x²

0 = -24*97200 + 8x²

Then

8x² = 24*97200

Dividing by sides by 8

x2 = 3*97200

Taking the positive square root of both sides (since negative distances make no sense in the real world):

x = √(3*97200) = 540

y = 97200/x = 97200/540 = 180

So the river side is 180 feet long and the side perpendicular to the river is 540 feet long.

Using the cost equation,

Cost = 24*180 + 540*8 = $8,640.

Rather than use the first derivative, as shown above, you could use a graphing calculator, such as a TI 83 or 84, input the cost function, and use it to find the x-value which provides the minimum value.

Answer 2

97200 =Lw

w =97200/L

C = 20(2L) + 4(2w) = 40L +8(97200)/L

take the derivative and set = 0

C' = 40 - 8(97200)/L^2 = 0

40L^2 - 8(97200)= 0

L^2- 2(9720) = 0

L^2= 2(9720)

L = square root of 2(9720) = about 139.43 feet Long parallel to the river

w = 97200/L = about 697.14 feet wide, perpendicular to the river

looks about right, as the L side cost 5 times as much, you might guess about 5 times shorter

use x and y instead of L and w if that's what they want