Express the area a of the rectangle as a function of x

Question

A rectangle has one corner on the graph of y=16-x^2, another at the origin, a third on the positive y axis and the fourth on the positive x axis

Answer 1

Express the area a of the rectangle as a function of x

A rectangle has one corner on the graph of y=16-x^2, another at the origin, a third on the positive y axis and the fourth on the positive x axis

Given:

A rectangle is bounded by x = 0, y = 0, and y = -x²+16 and has vertices somewhere on these equations and the vertex.

Let x be the width and y be the length of the rectangle so the equation will be A=xy.

If there's a corner on the origin, and somewhere on the y-axis, x-axis, and y = -x²+16, then y = -x²+16 can be used to dictate the area of the triangle. Thus, the area of the rectangle as a function of x is:

A(x) = xy = x(-x²+16) = -x³+16x