Respuesta :
Answer:
The required function is [tex]A(W)=-W^2+800W[/tex].
The value of W is 400 yd.
The maximum area is 160000 yd².
Step-by-step explanation:
Diana has available 1600 yards of fencing and wishes to enclose a rectangular area.
Let L be the length and W be the width of rectangle.
The perimeter of the rectangular area is 1600 yards.
i.e. [tex]P=2(L+W)[/tex]
[tex]1600=2(L+W)[/tex]
[tex]L+W=800[/tex]
[tex]L=800-W[/tex]
a) Express the area A of the rectangle as a function of the width W of the rectangle.
The area of the rectangle is [tex]A=L\times W[/tex]
[tex]A(W)=(800-W)\times W[/tex]
[tex]A(W)=-W^2+800W[/tex]
So, the required function is [tex]A(W)=-W^2+800W[/tex].
b) For what value of W is the area the largest ?
The area is largest when [tex]W=-\frac{b}{2a}[/tex]
From (a) the value of a=-1 and b=800
Substitute,
[tex]W=-\frac{800}{2(-1)}[/tex]
[tex]W=400[/tex]
So, the value of W is 400 yd.
c) What is the maximum area?
For maximum area substitute the value of W=400 in area,
[tex]A(400)=-400^2+800(400)[/tex]
[tex]A(400)=-160000+320000[/tex]
[tex]A(400)=160000[/tex]
So, the maximum area is 160000 yd².
