Inscribing a rectangle under a parabola?

Question

This problem stumped me, so I'm asking for help:

A rectangle is inscribed between the x-axis and the parabola (36-x^2) with one side along the x axis

find the maximum area of the rectangle and write a function of x for the area

Answer 1

The rectangle will be symmetric about the y-axis.

y = 36 - x²

The width will go from -x to x and therefore be 2x.
The height is y.
A = Area

A = 2xy = 2x(36 - x²) = 72x - 2x³

Take the derivative and set equal to zero to find critical values.

dA/dx = 72 - 6x² = 0
12 - x² = 0
x² = 12
x = √12 = 2√3

The solution must be positive since x is a distance.

A = 2x(36 - x²) = 2(2√3)(36 - 12) = 4√3(24) = 96√3

The maximum area is 96√3.

Answer 2

part of rectangle = width x top width = seventy two-0 = seventy two top = y = -x^2/36+2x section = seventy two [-x^2/36+2x] section = -72x^2/36 + 144x =-2x^2+144x dA/dx = -4x+one hundred forty four=0 4x=one hundred forty four x=one hundred forty four/4 = 36 section = seventy two [-x^2/36+2x] = seventy two [ (-36)^2/36 + 2(36)] = seventy two [ 36+seventy two ] = seventy two(108) = 7,776 squaregadgets