Consider the function. ƒ(x) = √(25 - x^2) on the closed interval [-5, 5]. The curve y = f(x) is drawn on the figure below (blue). A point (x, y) is on the curve.

Question

Consider the function.

ƒ(x) = √(25 - x²) on the closed interval [-5, 5].

The curve y = f(x) is drawn on the figure below (blue). A point (x, y) is on the curve.

The curve y = f(x) is the upper semi-circle of radius 5 with center at the origin.

A rectangle is inscribed in the semi-circle with the upper right corner at the point (x, y) and the upper left corner also on the semi-circle.

Write a function to compute the Area A of this rectangle as a function of x.

A =

For what number x is the area A maximized?

x =

Answer 1

The area of a rectangle is just the base times the height.

The base of the inscribed rectangle is 2x. The height of the rectangle is f(x)= √(25 - x²)

A=2x√(25 - x²)

Now, to maximize A, we take the derivative of A with respect to x, set it to zero, and then verify it is a maximum.

dA/dx 2x√(25 - x²) = 2(25 -2x²) / √(25 - x²)

Now set 2(25 - x²) /√(25 - x²) = 0 and solve

2(25 - 2x²) /√(25 - x²) =0

(25 - x²) =0

2x²=25

x²=25/2

x= ± 5√2/2

Since the leading coefficient in the quadratic is negative, we know we have a unique maximum., i.e., the function is concave.

If you evaluate both the negative and positive solutions for the zeros of the derivative, you will find that

x= +5√2/2 is the solution that maximizes the area