SOLUTION: David has 440 yards of fencing and wishes to enclose a rectangular area. 1. Express the area A of the rectangle as a function of the width W of the rectangle.
2. For what value of
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Question 629991: David has 440 yards of fencing and wishes to enclose a rectangular area. 1. Express the area A of the rectangle as a function of the width W of the rectangle.
2. For what value of W is the area largest?
3. What is the maximum area?
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
The perimeter, P = 2(l+w)
Expressing the length in terms of the width, we have
l = P/2 - w
The area, A = w*l = w(P/2 - w) = (P/2)w - w^2
The area is maximized when dA/dw = 0:
dA/dw = 0 = P/2 - 2w -> w = P/4, which means the area is maximized when l=w, ie. a square
Inserting the value for P gives w = 440/4 = 110 yd
A = 110*110 = 12,100 sq. yd.
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