Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola.
y=5-x^2

let 2x be the length of rectangw where x is value of x of point on parabola width is represented as y is the length.

  Area = 2x*y = 2x (5-x^2) = 10x -2x^3
 maximize Area by finding x value where derivative is zero
 dA/dx = 10 -6x^2 = 0
 --> x = sqrt(5/3)

optimal dimensions: length = 2sqrt(5/3) width = 10/3

I’m confused as to what you are asking 66 2/3 percentage wise