A Calculus student bought 20 m of flexible garden edging (shown in green). He plans to put two gardens in the back corners of his parents' property: one square and one in the shape of a quarter circle. He will use the edging on the interior edges (shown in green on the diagram).

Question

a) If x represents the section of edging used for the quarter circle, show that the total area for the two gardens can be modeled by the function below:
A(x)=x^2/π+(10−0.5x)^2.

*Image of a giant rectangular box with a green square in the top left corner and a green quarter circle in the top right*

b) How should the wise Calculus student split the edging into two pieces in order to maximize the total area of the two gardens? Remember, it could be entirely one of the two shapes.

Answer 1

So. You have a 1/4 circle of radius r with arc length x.
x = π/2 r
The area of the 1/4 circle is 1/4 πr^2 = 1/4 π(2x/π)^2 = x^2/π

The two sides of the square are each (20-x)/2 = 10-0.5x

so the area formula is correct. Now you just want to maximize the area. The derivative is
A'(x) = 2x/π+2(10−0.5x)(-0.5) = (2/π + 1/2)x - 10
This is zero when x = 20π/(π+4) ≈ 8.8

Answer 2

I am not sure what are interior edges but for the second part

dA/dx = 2 x/pi + 2 (10-.5x)(-.5)
= 2 x/pi -10 +.5 x
= (2/pi+.5)x-10
that is 0 for max or min of A
x = 10 / (2/pi+.5)