Find the maximum and minimum of $3x^2 +2xy$ inside and on the boundary of the square with vertices $(\pm1,\pm1)$.
I call the given expression $P$. I tried evaluating $P$ at the given vertices: $$P(1, 1) = 5 \\ P(-1, -1) = 5 \\ P(-1, 1) = 1 \\ P(1, -1) = 1 $$ So it looks like the maximum at the boundaries must be $5.$
I am not sure how to find the extrema of $P$ inside the square. One way I know of is to use Lagrange Multipliers, but I am not sure how to represent the square as an equation in $x, y.$
I have another similar problem:
Find the maximum and minimum of $x^2 −y^2 −2x+4y$ on and inside the triangle bounded below by the $x$-axis, above by the line $y = x + 2$ and on the right by the line $x = 2$.
Again, I call the given expression $P$. If $y = 0,$ then $x = -2$ and so one vertex is $(-2, 0).$ Also, $y = 4$ at $x = 2$ which gives $(2, 4).$ Another vertex is $(2, 0).$ Then $$P(-2, 0) = 8 \\ P(2, 0) = 0 \\ P(2, 4) = 0$$ So, at the boundaries of the given triangle $P$ has maximum $= 8$. Unsure how to find the extrema of $P$ as it passes through the inside the triangle. I think if I knew the level set against which $P$ must be optimized, I could possibly use the method of Lagrange Multipliers.
My question is How can I find the extrema (in particular, minima) in either problem? If it's done with the help of Lagrange Multipliers, how can I set up the necessary level sets? Thanks.
Here are some hints on your first problem.
To find the extrema of your expression in the interior of the square, do the same thing you would do for find extrema on the entire plane. Take the partial derivatives of your expression with respect to $x$ and to $y$, set them equal to zero, and solve the simultaneous equations. Any solution you find will be a critical point of the expression. Find those critical points that are inside the square, and for each one use the second partial derivatives or other means to determine if each is a local maximum, local minimum, or other.
Then there is one more possible location for extrema that you overlooked: the sides of the square. You could use Lagrange multipliers to find these extrema but I wouldn't bother. (If your teacher insists, you could use Lagrange multipliers, but I find the following method to be easier.) You could just parameterize each side in one variable, then use the common Calculus I methods to find the extrema for that one variable that lie inside the side (the line segment).
Note that your statement "it looks like the maximum at the boundaries must be 5" may well be false since you did not check the square's sides. You only know the maximum at the corners must be 5.
You know have three groups of extrema: at the corners, in the interior, and on the sides. Determine which of those is the absolute maximum and which is the absolute minimum, and you are done. (There are absolute extrema, since you have a continuous expression on a compact region.)