Finding the Maximum and Minimum Values of a Function in a Domain 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!
 A: As the guy in the comment says, the boundary of the domain is the circumference of the circle $x^2 + y^2 = 225$, not just the two points. Since the function has a minima at $(1,0)$ and no other critical points in the region, it means that the function is increasing in every direction from that point. So the maxima of the function will be at one of the boundary points. Now if the point is on the boundary, it will lie on the circle's circumference and satisfy the equation $x^2 + y^2 = 225$. Putting this in the function equation, we get
$$f(x,y) = 2x^2 + 3y^2 -4x -5$$
$$2x^2 + 3(225 - x^2) -4x -5$$
$$g(x)=-x^2 -4x + 670$$
Now you can diffrentiate to find out where this will have its maxima
$$g'(x) = -2x - 4 = 0$$
$$x=-2$$
As you can quickly confirm from the original equation or the graph of a quadratic equation, this is a maxima. Using this to get $y$
$$y^2 = 225 - x^2 = 225 -4 = 221$$
$$y = \sqrt{221}$$
Putting $x=-2$ and $y=\sqrt{221}$ back in the original function
$$f(x,y) = 674$$
A: 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.
A: 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.
A: With free CAS Maxima and my programm "nopt":
package "nopt"
examples

