A rectangular box has three of its faces on the coordinate planes and one vertex in the first octant on the paraboloid z=100-x2-y2. determine the maximum volume of the box.

The volume as a function of the location of that vertex is

... v (x, y, z) = x·y·z = x·y· (100-x²-y²)

This function is symmetrical in x and y, so will be a maximum when x=y. That is, you wish to maximize the function

... v (x) = x² (100 - 2x²) = 2x² (50-x²)

This is a quadratic in x² that has zeros at x²=0 and x²=50. It will have a maximum halfway between those zeros, at x²=25. That maximum volume is

... v (5) = 2·25· (50-25) = 1250

The maximum volume of the box is 1250 cubic units.