Let the constraint function . Note that the objective function is given by .
By the method of Lagrange multipliers, if is a local extremum of the function , which can be a candidate for the absolute extrema, on the ellipse, then it satisfies
for some
.
Since we have
a local extremum
on the ellipse should satisfy
The second equation implies that , so either or should hold.
In the case of , we get from the third equation. Then, from the first equation, we have , which is equivalent to and . Therefore, possible local extrema for the function on the ellipse are and .
On the other hand, if , then the first equation gives , so that . Then, for some the third equation can be satisfied---we don't need its value. Therefore, in this case, we obtain possible local extrema and .
Finally, we compare the function value of at theses candidates, to find the absolute extrema. Since
on the ellipse, the function
has its absolute maximum value
at
and
while it has its absolute minimum value
at
.
Answer: