Answer:
50 square feet.
Step-by-step explanation:
Let x represent the width of the fence and y represent the length of the fence.
One side of the length is already covered by the house. This leaves 2x and y. This must sum to 20. Hence:
[tex]2x+y=20[/tex]
The area of the enclosure will be given by:
[tex]A=xy[/tex]
From the first equation, we can subtract 2x from both sides:
[tex]y=20-2x[/tex]
Substitute:
[tex]A=x(20-2x)[/tex]
This is now a quadratic. Recall that the maximum value of a quadratic always occurs at its vertex. So, find the vertex of the equation. Distribute:
[tex]A=20x-2x^2[/tex]
In this case, a = -2, b = 20, and c = 0.
Hence, the vertex is:
[tex]\displaystyle x=-\frac{b}{2a}=-\frac{(20)}{2(-2)}=5[/tex]
Substitute this value back into the equation and evaluate:
[tex]A(5)=20(5)-2(5)^2=50\text{ ft}^2[/tex]
The maximum area is 50 square feet.
This occurs when the dimensions are 5 feet by 10 feet.
