What are the dimensions of the rectangular area that Louise roped off? And what is the maximum area of the rectangular area?

Question

Louise wants to build rectangular greenhouse on the side of her house. She starts of by roping off the area she wants to build in. Since one side of the greenhouse will be the side of her house, she only needs to rope off three sides. Louise uses 36 ft of rope and maximizes the enclosed area.

Answer 1

This is an optimization question from Calculus I and is solved as per the following:

Let L be the length of the area to be roped, and w the width of the area, the perimeter to be roped is therefore

L + 2w = 36 ft ( note that only one L is used since the other L is the side of the house) so lets solve for L:

L = 36- 2w ....eq 1 , where L is the constraint. now we need to maximize the area A = w L ...eq 2 for a rectangular area

Substitute eq 1 in 2, A(w)= w(36-2w) = 36w-2w² , we now find the derivative of the function

dA/dw = A'(w) = 36 - 4w setting this value equal to zero means that the slope of the function at point w is zero and will result in a critical point of the function. To test if this critical point is a maximum, the function must be decreasing on both sides of the point, Taking the 2nd derivative

A''(w) = -4, the value is always -ve and the function is decreasing on both sides of the critical point and the critical point is indeed a maximum

solving for w, 36 - 4w = 0, -4w = -36w , w = 9, with w =9, we solve for L in eq...1 , L + 2(9) = 36, L = 36-18 = 18 ft, and the Maximum area that can be generated is

A = (18ft) (9ft)= 162 ft²

Answer 2

36= 2W + L

L = 36-2W

A=LW = (36-2W)W = 36W -2W^2

A' = 36-4W = 0

4W = 36

W =36/4 = 9 feet

L =36-2(9) = 18 feet

Max area = 9(18) = 162 ft^2