Calculus- Early Transcendentals, 2021a

13.7. Maxima and Minima 477
Recall that when we did single variable global maximum and minimum problems, the easiest cases
were those for which the variable could be limited to a finite closed interval, for then we simply had to
check all critical values and the endpoints. The previous example is difficult because there is no finite
boundary to the domain of the problem—both w and l can be in (0,∞). As in the single variable case, the
problem is often simpler when there is a finite boundary.
Theorem 13.31: Multivariate Absolute Extrema
If f (x,y) is continuous on a closed and bounded subset of R 2 , then it has both a maximum and
minimum value.
As in the case of single variable functions, this means that the maximum and minimum values must
occur at a critical point or on the boundary; in the two variable case, however, the boundary is a curve, not
merely two endpoints.
Example 13.32: Optimizing Volume of a Box
The length of the diagonal of a box is to be 1 meter; find the maximum possible volume.
Solution. If the box is placed with one corner at the origin, and sides along the axes, the length of the
diagonal is √ x 2 + y 2 + z 2 , and the volume is
V = xyz = xy √ 1 − x 2 − y 2 .
Clearly, x 2 + y 2 ≤ 1, so the domain we are interested in is the quarter of the unit disk in the first quadrant.
Computing derivatives:
V x = y − 2yx2 − y 3
√
1 − x 2 − y 2
V y = x − 2xy2 − x 3
√
1 − x 2 − y 2
If these are both 0, then x = 0ory = 0, or x = y = 1/ √ 3. The boundary of the domain is composed of three
curves: x = 0fory ∈ [0,1]; y = 0forx ∈ [0,1];andx 2 + y 2 = 1, where x ≥ 0andy ≥ 0. In all three cases,
the volume xy √ 1 − x 2 − y 2 is 0, so the maximum occurs at the only critical point (1/ √ 3,1/ √ 3,1/ √ 3).
See Figure 13.10.
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