A rectangular area is to be enclosed and divided into thirds. The family has $400 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? (The answer contains a fraction.)

The dimensions that will maximize the area enclosed is 100/3 feet sq.

Explanation:

A rectangle has a width and a length.

Taking W as Width and L as Length

Area = W x L

Draw a rectangle and divide it into 2/3 with two lines parallel with the width.

The outside fence perimeter ia given as 2 W + 2 L, while the inside fence length will be 2 W

Therefore, the total cost is gotten as:

[2 W + 2 L] x $10 + 2 W x $20 = $400

Open the bracket,

= 20 W + 20 L + 40 W = $400

= 60 W + 20 L = 400

Make L the subject of formula by dividing both sides by 20,

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

= 3 W + L = 20 Equation A

In order to make L the subject of formular still, you will subtract 3 W from both sides of equation A,

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

L = 20 - 3W Equation B

Since Area of a rectangle = W x L, and L = 20 - 3W. Then substitute 20 - 3W as L in the equation below,

= W x ( 20 - 3W )

Open the bracket,

= 20 W - 3 W²

1st derivative:

dA / dW = 20 - 3 x 2 x W

= 20 - 6 W

Assume dA / dW = 0

20 - 6 W = 0

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

20 = 6 W

Make W the subject of formula

W = 20/6

We can rewrite it as

W = 2 x 10 / ( 2 x 3 )

So W = 10 / 3

W = 10 / 3 feet

NOTE:

A function has a local maximum or minimum if 1st derivative = 0.

If 2nd derivative < 0 then function has a maximum, but if 2nd derivative > 0 then function has a mimum.

However, 2nd derivative = -6 here, which means function has a maximum.

For W = 10 / 3 feet,

L = 20 - 3W

= 20 - 3 x 10 / 3

= 20 - 10

= 10 feet

The dimensions that will maximize the area enclosed = the Maximum,

It is gotten as:

W x L = 10 / 3 x 10

= 100 / 3 feet sq.

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