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David has available 480 yards of fencing and wishes to enclose a rectangular area.

David has available 480 yards of fencing and wishes to enclose a rectangular area-example-1

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Answer:

Explanation:

Total fencing David has = 480 yards

Means He can surround all 4 sides of the rectangular area using this fencing.

We can say that the total fencing he has in terms of length L and width W of the rectangle will be L+L+W+W = 480

So , 2(L+W) =480

Divide both sides by 2 to get

L+W = 240

Subtract W from both sides so we get L in terms of W

L= 240 - W

(a) We know that the area of a rectangle is Length times width (L Ă— W)

So , Area = (240-W) W

Area = 240W - W^2

(b) By comparing the area equation with ax^2+bx+c (standard quadratic equation)

we get a=-1 and b =240

Area is maximum when


W = (-b)/(2a) \\W= (-240)/(2(-1)) \\W = 120

So for W = 120 yd the area is largest .

(c) Maximum area of the rectangle can be found by plugging this W = 120 in the Area equation that we got in part (a)

So the maximum area ,

A =
240 (120) - 120 ^2

A =28800-14400

A = 14400

Hence Maximum area = 14400 sq yd

David has available 480 yards of fencing and wishes to enclose a rectangular area-example-1
User Hagay
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