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!
Best Answer
According to the method of Lagrange parameters, the gradient of $f$ will be normal to the boundary in an extremal value. The normal to the circle in a given point $(x,y)$ is just the position vector $(x,y)$, so you have to look for points where $\nabla f$ is a mulitple of that vector.
Now
$$\nabla f =\left ( \array{4x-4 \\ 6y} \right) $$ which means we are looking for $x, y$ such that $ 4x-4 = \lambda x, 6y= \lambda y$ with the side condition $x^2+y^2=225$
This is true iff ($\lambda = 6$ or $y=0$).
If $\lambda = 6$ you easily see that then $x=-2$ and $y$ is determined by the side condition.
If $y=0$ $x$ is determined by the side condition (and $\lambda$ by the equation for $x$ and $\lambda$).
Plugging in $x$ and $y$ into the function will show whether you have a maximum or minimum.