A homeowner wants to enclose a rectangular piece of land beside a river. He must put a fence on 3 sides of the field (the river forms the fourth side). The field has length L (which runs parallel to the river) and width W. The homeowner has 200 yards of fencing material. What should the length of the field L and the width of the field W be in order to produce the maximum possible area using all 200 yards of fence?

Question

A homeowner wants to enclose a rectangular piece of land beside a river. He must put a fence on 3 sides of the field (the river forms the fourth side). The field has length L (which runs parallel to the river) and width W. The homeowner has 200 yards of fencing material. What should the length of the field L and the width of the field W be in order to produce the maximum possible area using all 200 yards of fence?

So I know that A=WL, and that 2W+L=200
I changed L to 200-2W, then got
A=W(200-2W)=200W-2W^2

But i'm not sure what to do next, I've been searching and would be really grateful if someone can help, thanks!

Answer 1

the max A lies on the axis of symmetry

Wmax = -b / 2a = -200 / (2 * -2) = 50

Answer 2

To find the dimensions of the field that would produce the maximum possible area using all 200 yards of fence, you need to maximize the area equation A = WL, subject to the constraint 2W + L = 200.

You already correctly simplified the constraint equation to L = 200 - 2W. Now substitute this expression into the area equation:

A = W(200 - 2W) = 200W - 2W^2

To find the maximum area, you need to take the derivative of A with respect to W and set it equal to zero:

dA/dW = 200 - 4W

Setting this equal to zero:

200 - 4W = 0
4W = 200
W = 50

Now that you have the value of W, substitute it back into the constraint equation to find the corresponding value of L:

2W + L = 200
2(50) + L = 200
100 + L = 200
L = 100

Therefore, the width of the field should be 50 yards and the length should be 100 yards in order to produce the maximum possible area using all 200 yards of fence.