Find the area of the largest rectangle with one corner on th

Find the area of the largest rectangle with one corner on the origin, opposite corner on the parabola f(x)=-x^2+12, and sides parallel to the axes

So Assuming Opposite Corner is at (x,y)

Then

y = x² +12

Since sides are parallel to axes

Now Length = x

Height = y = x² + 12

Area

A = x*(x² +12)

dA/dx = (x²+ 12) + x*(2x) = 3x² + 12

Which is alays +ve

So no maximum value Since it can be