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.
To find the total area of the two gardens, we need to find the areas of the square garden and the quarter circle garden, then add them together.
a) Let's start with finding the area of the square garden. Since the length of each side of the square garden is given by (10 - 0.5x) meters (as shown in the diagram), the area of the square garden is (10 - 0.5x)^2 square meters.
Next, let's find the area of the quarter circle garden. The length of the curved part of the quarter circle is given by x meters (as shown in the diagram). The formula to calculate the area of a quarter circle is A = x^2/π (where A is the area and x is the radius).
So, the total area for the two gardens is given by the function:
A(x) = x^2/π + (10 - 0.5x)^2 square meters.
b) To find how the Calculus student should split the edging into two pieces to maximize the total area, we need to find the maximum value of the function A(x).
To find the maximum, we can take the derivative of the function A(x) with respect to x and set it equal to zero. So, let's differentiate the function A(x) with respect to x:
A'(x) = (2x/π) - 2(10 - 0.5x)(0.5)
Now, set A'(x) equal to zero and solve for x:
(2x/π) - (10 - 0.5x) = 0
2x/π - 10 + 0.5x = 0
(2x + 0.5πx)/π - 10 = 0
(2.5x + 0.5πx)/π - 10 = 0
To solve this equation, we can multiply through by π to remove it from the denominator:
2.5x + 0.5πx - 10π = 0
Now, we can solve this equation for x to find the value that maximizes the total area.
apparently the side of the square is 10-.5x
(a) the 1/4 circumference of the circle is x, so the radius r can be found by
2πr/4 = x
r = 2x/π
So, the area of the 1/4 circle is 1/4 πr^2 = π/4 (2x/π)^2 = x^2/π
(b) to maximize the area, set its derivative to zero:
A(x)=x^2/π+(10−0.5x)^2 = (1/π + 1/4)x^2 - x + 100
This is a parabola which opens up, so it has no maximum, except at one of the ends of the interval. That means either the square or the circle has area zero.
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