Richard P. answered • 06/22/14

Fairfax County Tutor for HS Math and Science

This problem can be set up as an optimization problem in calculus

The points on the parabolic curve can be written as (x, 11- x

^{2})so the upper right corner will have the coordinates of (x, 11-x

^{2}) for some optimal value of xThe width = 2x and the height = 11- x

^{2}again for some optimal value of xThe objective function is the area (which is to me maximized)

area = width x height = A = 2x (11-x

^{2})The calculus approach is to compute the derivative of A with respect to x and set this equal to zero

The analytic derivative of A is 22 - 6 x

^{2}setting this equal to zero yields x = sqrt(11/3) = 1.914 , substituting this result back into the expressions for width and height gives

width = 2 sqrt(11/3) and height = 22/3

The graphing calculator approach is to put the function 2x (11 - x

^{2}) into Y1 and plot the curve.Then use 2nd calc ==> maximum

The result is x = 1.914