Find the absolute minimum and maximum of $f(x,y)=x^2y+2xy+12y^2$ on the ellipse $x^2+2x+16y^2\leq{8}$

Question

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.

Answer 1

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

Answer 2

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.

1. $y\geq0.$

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 $y\geq0$.

2. $y\leq0$.

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 $$\min\limits_{x^2+2x+16y^2\leq8}(x^2y+2xy+12y^2)=\frac{5}{2}-\frac{11}{6}\sqrt{\frac{11}{3}}.$$