What are the dimensions of such a rectangle with the greatest possible area?

Question

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=12-x² . What are the dimensions of such a rectangle with the greatest possible area?

width=?

height=?

 

Answer 1

The parabola opens down ward  and the vertex is located at (0,12),  The focal distance is p which can be calculated as y = 11.75 Then the largest distance of the cord will be the two x intersects and this length is given by the equation x=(12-Y)^0.5 which has two points one negative and other positive, the length of the base will be (x-(-x)) =2x = 2*(12-Y)^0.5

The rectangle area will be= Y*(2*(12-Y)^0.5) = 2Y*(12-Y)^0.5

to maximize I can take the first derivative and equal it to zero or use Solver to obtain the numbers, I picked the second.

I obtained Y =8 and area equals 32; the dimensions are:

height = 8

width = 2*2=4

Check!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!