Find the absolute minimum and maximum of $f(x,y)=x^2y+2xy+12y^2$ on the ellipse $x^2+2x+16y^2\leq{8}$ I want to find the absolute minimum and maximum of $f(x,y)=x^2y+2xy+12y^2$ on the ellipse $x^2+2x+16y^2\leq{8}$
To calculate the critical points of $f$ I use the partial derivatives $f_x=2xy+2y$ and $f_y=x^2+2x+24y$ and the critical points will be those which $2xy+2y=0$ and $x^2+2x+24y=0$. 
When I try to solve this two-equation system I don't get a point like $(a,b)$, I get the points $(0,y),(0,0),(-2,0)$ This, I guess means that the points $(0,0),(-2,0)$ and all the points of the segment from $(0,-y)$ to $(0,y)$ for $|y|\leq{\sqrt{\frac{1}{2}}}$ are critical points. Since $f'(0,0)=0$ and $f'(-2,0)=0, (0,0)$ and $(-2,0)$ are both the absolute minimums of the function. And the absolute maximum would be $(0,\frac{1}{\sqrt{2}})$ and $(0,\frac{-1}{\sqrt{2}})$ (because $f(0,y)=12y^2)$
Is this correct?
Do I still need to calculate the critical points on the boundary?
Thank you for your time.
 A: I have used Lagragian Multiplier's method to find the maximum and minimum of f(x,y).
$f(x,y) = f(x,y)=x^2y+2xy+12y^2$
$g_1(x,y) = x^2+2x+16y^2\leq{8}$
$\nabla f = \lambda \nabla g_1$
$(2xy+2y)\hat i +(x^2+2x+24y)\hat j = \lambda\left( (2x+2)\hat i + 32y\hat j\right)$
$(x+1)y = \lambda(x+1)$
$(y-\lambda) (x+1) = 0\tag 1$
$x^2+2x+24y = 32\lambda y\tag 2$
From (1), it is either $x = -1$ or $\lambda = y$.
Case 1:
Thus $x = -1$
Now plug the value of $x$ in $g_1(x,y)$
you get $y_1 =  \frac{3}{4}$ and $y_2 = - \frac{3}{4}$ 
Now you have two points and $(x,y_1)$ and $(x, y_2)$.  Substitute these value in the f(x,y) and get the maximum and minimum.
Case 2:
If $\lambda = y$,
$x^2+2x+24y=32y^2$ and thus $x^2+2x = 32y^2-24y$
substitute this expression for $x^2+2x$ in $g_1(x,y)$ and you will get a quadratic and solve for y. and thus you will have two y's and and for each y you will have two x's and you will get four points and evaluate them and find the minimum.
Good luck
A: If $x=-1$ and $y=-\frac{3}{4}$ then we get a value $\frac{15}{2}$.
We'll prove that it's a maximal value.
Indeed, by the given $$(x+1)^2+16y^2\leq9.$$
Thus, $$x^2y+2xy+12y^2=y(x+1)^2+12y^2-y\leq$$
$$\leq|y|(x+1)^2+12y^2-y\leq|y|(9-16y^2)+12y^2-y.$$
Hence, it remains to prove that $$|y|(9-16y^2)+12y^2-y\leq\frac{15}{2}.$$
Consider two cases.


*

*$y\geq0.$


Thus, $0\leq y\leq\frac{3}{4}$ and we need to prove that 
$$y(9-16y^2)+12y^2-y\leq\frac{15}{2}$$ or
$$(4y+3)(8y^2-12y+5)\geq0,$$
which is obvious even for all $y\geq0$.


*$y\leq0$.


Thus, $-\frac{3}{4}\leq y\leq0$ and we need to prove that 
$$-y(9-16y^2)+12y^2-y\leq\frac{15}{2}$$ or
$$(4y+3)(8y^2-5)\leq0,$$
which is true for $-\frac{3}{4}\leq y\leq0$.
Now, about the minim.
Let $y\geq0$.
Thus, $$x^2y+2xy+12y^2=y(x+1)^2+12y^2-y\geq12y^2-y=12\left(y-\frac{1}{24}\right)^2-\frac{1}{48}\geq-\frac{1}{48}.$$
Let $y\leq0$.
Thus, $-\frac{3}{4}\leq y\leq0$ and
$$x^2y+2xy+12y^2=y(x+1)^2+12y^2-y\geq y(9-16y^2)+12y^2-y=$$
$$=-16y^3+12y^2+8y\geq\frac{5}{2}-\frac{11}{6}\sqrt{\frac{11}{3}},$$
where the equality occurs for $y=\frac{1}{4}-\frac{1}{4}\sqrt{\frac{11}{3}}.$
Since $$-\frac{1}{48}>\frac{5}{2}-\frac{11}{6}\sqrt{\frac{11}{3}},$$ we see that
$$\min\limits_{x^2+2x+16y^2\leq8}(x^2y+2xy+12y^2)=\frac{5}{2}-\frac{11}{6}\sqrt{\frac{11}{3}}.$$
