TITLE: Finding the Maximum and Minimum Values of a Function in a Domain
QUESTION [5 upvotes]: Find the maximum and minimum values of the function $f(x, y) = 2x^2+3y^2-4x-5$ on the domain $x^2+y^2\le 225$.
After finding the first partial derivatives, I found that $(1, 0)$ was a critical point and I found that it was a local minimum from the second derivative test. So the minimum value of $f(x, y)$ would be $-7$ at the point $(1, 0)$.
However, what I am confused about is how to find the maximum value and point. Since this function does not have a local maximum point, I thought that the answer would simply be on the boundaries of the inequality. However, it seems that neither the point $(15, 0)$ or $(0, 15)$ give the correct answer.
If anyone knows how I should approach this problem and can provide feedback, I would be very grateful!

REPLY [2 votes]: If you don't know, or want to use, Lagrange Multipliers as in Thomas's answer, and you want to find the local extrema on the boundary to make sure you don't miss anything, then you can either handle the endpoints with $x=\pm15$ that Sauhard Sharma neglected to address in their answer, or you can give the circle a more natural parametrization.
Circles can be parametrized with cosines and sines, so the circle $x^2+y^2=225$ is parametrized by $x=15\cos t,\, y=15\sin t$. If you substitute this in to the objective function $2x^2+3y^2-4x-5$, you'll see how the values of the function above the circle depend on the angle $t$, and you can then use single-variable calculus to find the local extrema. Since cosine and sine are continuous on the whole real line, you don't have to worry about missing end-points with this method.