Two variable function limited in a triangle

Question

Given the function $ 2x^2+5y^2-1 $, how do I find its maxima/minima in the triangle $ x\geq0, y\geq 0, 3y+x \leq 2 $, which has its vertices in $ (0,0),(2,0),(0,2/3) $?

Answer 1

Call $T$ your domain, i.e. the triangle, and $f= f(x,y)$ your function. If $(x,y)$ in the interior of $T$ is a point of maximum or minimum, then the gradient of $f$ vanishes at it. You can easily see that $\frac{\partial f}{\partial x}(x,y) = 4x = 0$ and $\frac{\partial f}{\partial y}(x,y) = 10y = 0$ means $(x,y) = (0,0)$, which is not in the interior of $T$. So there are no maxima and minima in the interior of $T$.

Therefore, you should look for maxima and minima on the boundary. If $x = 0$ then $f(0,y) = 5y^2-1$, and you have then the maximum when $y=\frac{2}{3}$, and the minimum when $y = 0$.

Do the same when $y = 0$.

The last segment of the boundary of $T$ is the line segment given by $y = -\frac{x}{3}+\frac{2}{3}$. In this case $f(x,y) = f(x,-x/3+2/3)$, and so you get a function of one variable. You can study its maximum and minimum when $x$ ranges between 0 and 2.

Finally, compare all the results found to get maxima and minima of $f$.