Question
Find the maximum area of a rectangle that lies above the x axis and below y=9-x^2
Answers
Steve
any candidate rectangle must have its vertices on the x-axis and on the curve.
If (x,y) is a vertex, so is (-x,y)
A = 2x(9-x^2) = 18x - 2x^3
dA/dx = 18 - 6x^2
A' = 0 when x = √3
A(√3) = 2√3(9-3) = 12√3
If (x,y) is a vertex, so is (-x,y)
A = 2x(9-x^2) = 18x - 2x^3
dA/dx = 18 - 6x^2
A' = 0 when x = √3
A(√3) = 2√3(9-3) = 12√3