Pamela hired a contractor to paint 75% of
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Lagrange multipliers:
(if 
)
(if )
(if 
)
In the first octant, we assume
, so we can ignore the caveats above. Now,
so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of
.
We also need to check the boundary of the region, i.e. the intersection of
with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force 
, so the point we found is the only extremum.
In the first octant, we assume
so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of
We also need to check the boundary of the region, i.e. the intersection of