Barbara (her friends call her Barb because she solves problems about wire) is planning a sustainable fish farm venture for her large property. She has a total of 558 meters of wire for the electrified fencing which will be used to enclose two unconnected pool areas. One of the areas must be square, while the other must be a rectangle that is two times as long as it is wide. The square pond enclosure must contain at least 144 square meters and the other one (the rectangle) at least 1152 square meters. Find the maximum total area of the two enclosures.

Question

Barbara (her friends call her Barb because she solves problems about wire) is planning a sustainable fish farm venture for her large property. She has a total of 558 meters of wire for the electrified fencing which will be used to enclose two unconnected pool areas. One of the areas must be square, while the other must be a rectangle that is two times as long as it is wide. The square pond enclosure must contain at least 144 square meters and the other one (the rectangle) at least 1152 square meters. Find the maximum total area of the two enclosures.

I got A=x^2+l*w --> A=x^2+2w^2

Then I use the perimeter equation
558=4x+2w+2l --> 558 = 4x + 6w
w = 93 - 2/3x

Then I plugged it into the area equation
A= x^2+2(2/3x)^2

Then I found the derivative
A(prime) = 34/9x-248

what do I do next?

Answer 1

I think you went astray here:

A= x^2+2(2/3x)^2
It should be
A= x^2+2w^2 = x^2 + 2(93 - 2/3 x)^2 = 1/9 (17x^2 - 2232x + 155682
dA/dx = 1/9 (34x - 2232)
A is a minimum when dA/dx = 0 (x = 65), but that is not what we want.
Instead, check the value of A at the edges of the domain, to find the absolute maximum.
Since 2w^2 >= 1152, w >= 24
since x^2 >= 144, x >= 12
So, using the minimum w (and thus the maximum x)
A = 1152 + ((558-6w)/4)^2 = 11864.25
Using the minimum x,
A = 144 + 2(93 - 2/3 x)^2 = 14594
So that is the maximum area.