Question

What is the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve y=12-x^2?

Answer 1

Let
\[
A(x) = 2x(12-x^2)= 24 x - 2x^3\\
A'(x) = 24 - 6 x^2=6( 4-x^2)=6(2-x)(2+x)
\]
You can verify that A(x) is maximum at x=2
and A(2)=32.
The answer is 32