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.

  • $\begingroup$ $f_x(0,y)=2y$ and $f_y(0,y)=24y$. How are these zero for all $|y|\le\sqrt{1/2}$? Moreover, how does the fact that $f$ is zero at two of your critical points lead you to conclude that these are minima? $\endgroup$
    – amd
    Oct 24 '17 at 23:56

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

  • $\begingroup$ After reading and studying it online I see this method is the most common one to find these locals and minimums of multi variable functions in a domain. I understand your answer and I think the maximum is right, however when $\lambda = y$, $(2)$ can be true. The equality $(2)$ will happen when, for example, when $x=0,y=0$. Therefore, if $y=\lambda$, how would the problem continue? Thank you for your time $\endgroup$ Oct 25 '17 at 20:24
  • $\begingroup$ You are right sir, See edited answer $\endgroup$ Oct 26 '17 at 1:09

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 $y\geq0$.

  1. $y\leq0$.

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

  • $\begingroup$ How did you find the point $(-1,\frac{-3}{4})$? $\endgroup$ Oct 25 '17 at 16:57
  • $\begingroup$ @John Keeper When we get the function of the variable $y$ the rest is smooth. $\endgroup$ Oct 25 '17 at 18:04
  • $\begingroup$ I don't understand the motivation behind $x^2y+2xy+12y^2=y(x+1)^2+12y^2-y$, and then trying to prove a inequality, why do you do it? $\endgroup$ Oct 25 '17 at 20:49
  • $\begingroup$ @John Keeper The condition gives $(x+1)^2+16y^2\leq9$ or $(x+1)^2\leq9-16y^2$ and we obtain a possibility to get a function of one variable $y$. $\endgroup$ Oct 26 '17 at 2:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.