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A rectangle is inscribed with its base on the x-axis and its upper corners
on the parabola y=9−x2. What are the dimensions of such a rectangle
with the greatest possible area?
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The area of any such inscribed rectangle is
A = A(x) = base*height = (2x)*y = 2x*(9-x^2) = 18x - 2x^3.
We want maximize the area A(x), i.e. maximize function A(x).
So, we take the derivative and equate it to zero
A'(x) = 18 - 6x^2 = 0.
From this equation, we find
6x^2 = 18 ---> x^2 = 18/6 = 3 ---> x = .
The dimensions of the inscribed rectangle are horizontally
and = 9 - 3 = 6 units vertically.
The maximum area is xy = = square units.
Solved.