Posted by Patrick on .
Find the dimensions of the rectangle with the largest area that is inscribed inside the parabola y = 16- x^2 and the x-axis
since we want a rectangle, the top side to be parallel to the bottom side. So, the rectangle is centered over (0,0).
Let the base of the rectangle extend from -x to x
The area is thus
a = 2xy = 2x(16-x^2)
da/dx = 32 - 6x^2
da/dx=0 when x = 4/√3
the rectangle is thus 8/√3 by 32/3