Answer: 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.
