.
Its dimensions are 2x, 2y, and 2z. In addition z is a function of x and y so
the volume V(x,y) is given
by
.
Computing the partials,
Next, from the geometry of the problem, if x = 0 or y = 0, then the volume
of the box is 0.
So at a maximal point,
To see that there is a maximum in the first quadrant at
,
let C be the region in
the first quadrant by the line x = 0, the line y = 0 and the ellipse
.
Then:
1.
on the boundary of C.
2.
for (x,y) inside C.
3.
space
is a continuous function.
So
must have a maximum inside C.
Our analysis of critical points shows that
is the only critical point in the
first quadrant. Consequently, the function
does indeed have a max at this critical point.
(An alternative argument to show that the function has a max at the critical
point could be based on
the second derivative test. This would mean
calculating the second order partials)
The maximal value is
.