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).
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.
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
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)
To solve this problem, we need to break it down into smaller steps.
a) First, let's find the area of each garden individually and then add them together to get the total area.
- The area of the square garden is given by one side squared, which we can call x (since it represents a section of the edging used for the quarter circle):
Area of square garden = x^2.
- The area of the quarter circle garden can be found by calculating half the circumference of the quarter circle and multiplying it by the radius (which is 10 - 0.5x). Since the circumference of a full circle is 2πr, the circumference of the quarter circle is πr/2:
Area of quarter circle garden = (πr/2) * r = (π/2)(10 - 0.5x)^2.
Now, we just need to add these two areas together to get the total area:
Total area = Area of square garden + Area of quarter circle garden
= x^2 + (π/2)(10 - 0.5x)^2.
So, the function that models the total area of the two gardens is given by:
A(x) = x^2 + (π/2)(10 - 0.5x)^2.
b) To determine how the Calculus student should split the edging in order to maximize the total area, we need to find the maximum point of the function A(x).
To find the maximum of a function, we can take the derivative and set it equal to 0. Let's differentiate A(x) with respect to x:
A'(x) = 2x + (π/2) * 2(10 - 0.5x)(-0.5) (using the power rule for derivatives).
Setting A'(x) equal to 0, we have:
2x + (π/2) * 2(10 - 0.5x)(-0.5) = 0.
Now, we can solve this equation to find the value of x that maximizes the total area.