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Use Lagrange multipliers to find the extrema of $f(x,y,z)=yz+x^2$ on the boundary of $J$ given by $\partial J=\{(x,y,z) \in \mathbb{R}^3 : x^2+2y^2+3z^2=6\}$ .

I have started by forming $L(x,y,z,\lambda) = f(x,y,z) - \lambda g(x,y,z)$, where $g(x,y,z)$ is the boundary condition. Here I have the gradient of $L(x,y,z,\lambda) = (2x-2x\lambda, z-4y\lambda, y-6z\lambda, -x^2-2y^2-3z^2+6)$. Making this equal to $0$ I got $4$ equations:

  1. $2x-2x\lambda=0$
  2. $z-4y\lambda=0$
  3. $y-6z\lambda=0$
  4. $-x^2-2y^2-3z^2+6=0$

This is where I am a little stuck. My first thought would be to use 1. so that $2x(1-\lambda)=0$, so either $x=0$ or $\lambda=1$. By using $x=0$ I would rewrite 4 and solve by substituting in values of $y$ and $z$ from 2 and 3 into 4. And if $\lambda=1$, I would have 3 different values of y=z - is this correct? i.e. I would have $z=4y$ from 2, $y=6z$ from 3 and $y=7/5z$ from making 2 equal 3. Am I on the right track?

Thanks.

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  • $\begingroup$ Here is a MathJax Tutorial: meta.math.stackexchange.com/questions/5020/…. It will make your question much easier to read. I think you are on the right track. When $\lambda=1$, you can then conclude that $y=z=0$. $\endgroup$ – KittyL Mar 5 '15 at 11:04
  • $\begingroup$ Thanks for the link and the help - i'll try to make any future questions with MathJax $\endgroup$ – Orangegulf Mar 5 '15 at 11:23
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Yes. I think you are on the right track. From Equation 1, it's easy to find $x=0$ or $\lambda=1$. Now we need to discuss it from two aspects.

(1) If $x=0$, we need to solve the three equations: \begin{equation} \begin{cases} &z-4y\lambda=0\\ &y-6z\lambda=0\\ &2y^2+3z^2-6=0 \end{cases} \end{equation} Hence we can get,

Case One: $\lambda=\frac{1}{2\sqrt{6}},y=\sqrt{\frac{3}{2}},z=1$ or $y=-\sqrt{\frac{3}{2}},z=-1$. Then $f(x,y,z)=\sqrt{\frac{3}{2}}$.

Case Two: $\lambda=-\frac{1}{2\sqrt{6}},y=-\sqrt{\frac{3}{2}},z=1$ or $y=\sqrt{\frac{3}{2}},z=-1$. Then $f(x,y,z)=-\sqrt{\frac{3}{2}}$.

(2)If $\lambda=1$, then we need to solve \begin{equation} \begin{cases} &z-4y=0\\ &y-6z=0\\ &x^2+2y^2+3z^2-6=0 \end{cases} \end{equation} Here $y=z=0,x^2=6$. Thus $f(x,y,z)=6$.

Based on (1) and (2), we can get the results:

$\min_{(x,y,z)\in J}f(x,y,z)=-\sqrt{\frac{3}{2}}$

$\max_{(x,y,z)\in J}f(x,y,z)=6$

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  • $\begingroup$ Excellent glad I was on the right track. Thanks a lot for your help! $\endgroup$ – Orangegulf Mar 5 '15 at 14:43

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