Question

A rectangular area is to be enclosed and divided into thirds. The family has $880 to spend for the fencing material. The outside fence costs $10 per foot installed, and the dividers cost $20 per foot installed. What are the dimensions that will maximize the area enclosed

Answer 1

\W = Width

L = Length

Draw rectangle W x L

Divde rectangle into thirds with two lines parallel with width.

Outside fance perimeter = 2 W + 2 L

Inside fance length = 2 W

Total cost :

( 2 W + 2 L ) * 10 $ + 2 W * 20 $ = 400 $

( 2 W + 2 L ) * 10 + 2 W * 20 = 400

20 W + 20 L + 40 W = 400

60 W + 20 L = 400 Divide both sides by 20

60 W / 20 + 20 L / 20 = 400 / 20

3 W + L = 20 Subtract 3 W to both sides

3 W + L - 3 W = 20 - 3 W

L = 20 - 3W

Area :

A = W * L =

W * ( 20 - 3W ) =

20 W - 3 W ^ 2

First derivation :

dA / dW = 20 - 3 * 2 * W = 20 - 6 W

If first derivation = 0 a function has a local maximum or a local minimum.

dA / dW = 0

20 - 6 W = 0 Add 6 W to both sides

20 - 6 W + 6 W = 0 + 6 W

20 = 6 W Divide both sides by 6

20 / 6 = 6 W / 6

20 / 6 = W

2 * 10 / ( 2 * 3 ) = W

10 / 3 = W

W = 10 / 3 ft

If second derivative < 0 then function has a maximum.

If second derivative > 0 then function has a mimum.

In this case second derivative = - 6

Second derivative < 0 so function has a maximum.

For W = 10 / 3 ft

L = 20 - 3W = 20 - 3 * 10 / 3 = 20 - 10 = 10 ft

Amax = W * L = 10 / 3 * 10 = 100 / 3 ft ^2