A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=11-x^2. What are the dimensions of such a rectangle with the greatest possible area?

Question

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=11-x^2. What are the dimensions of such a rectangle with the greatest possible area?

Answer 1

What a ridiculous answer by the bot

The max height of the parabola is 11, so how can the rectangle have a height of 11 ??

let the upper corner be at(x, 11-x^2)
A = 2x(11-x^2) = 22x - 2x^3
dA/dx = 22 - 6x^2 = 0 for a max of A
x^2 = 22/6 = 11/3
x = appr 1.91, then y = 22/3

Answer 2

The dimensions of the rectangle with the greatest possible area are 4 units by 11 units. The base of the rectangle is on the x-axis, so the width of the rectangle is 4 units. The upper corners of the rectangle are on the parabola y=11-x^2, so the height of the rectangle is 11 units.