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Find the Max and Min of $f(x,y,z)=y^2+4z^2-4yz-2xz-2xy$ under the constraint $g(x,y,z)=2x^2+3y^2+6z^2-1=0$

My work:

I used Lagrange theorem to get the following equation system:

$-2z-2y=\lambda 4x$

$2y-4z-x=\lambda 6y$

$8z-4y-2x=\lambda 12z$

By dividing by 2 and multiplying the first equation by $x$, the second by $y$ and the third by $z$, and then summing them all up we get that $\lambda(2x^2+3y^2+6z^2)=\lambda=y^2+4z^2-4yz-2xz-2xy=f(x,y,z)$

I'm not sure how to carry on from here, can anyone help?

1 Answers 1

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So, when solving a problem through Lagrange's undetermined multiplier, you have 2 pieces that end up being considered. Your first 3 equations determine the relationship between the undetermined multiplier and your variables (x,y,z), your last is the constraint g(x,y,z)=0.

The first 3 equations you have are the derivatives and allow you to find critical values in terms of $\lambda$: $$solution= (x_{critical}(\lambda),y_{critical}(\lambda),z_{critical}(\lambda))$$ Note, you can easily have more than 1 solution.

With a given constraint function $g(x,y,z)=0$, we want to find the constraint evaluated at the critical values $g(x_{critical}(\lambda)_{critical},y(\lambda),z_{critical}(\lambda))=0\rightarrow g(\lambda)=0$.

Solving for the roots here allows you to plug $\lambda$ back into the critical values to calculate the extrema.

If you wanted to solve your first 3 equations I would save some time and use matrices: $$4\lambda x+-2y-2z=0$$ $$x+(6\lambda -2)y+4z=0$$ $$-2x-4y+(8-2\lambda)z=0$$ Which can be written as: $$\begin{bmatrix}4\lambda & -2 &-2 \\1&(6\lambda-2) &4\\-2 &-4&(8-2\lambda)\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0 \\0\\0\end{bmatrix}$$ In more general terms, our system of equation is represented as: $$\textbf{M} \vec{v}=\vec{a}$$ $$\vec{v}_{critical}=\textbf{M}^{-1}\vec{a}$$ Leaving us with: $$\begin{bmatrix}x_{critical}\\y_{critical}\\z{critical}\end{bmatrix}=\frac{1}{det\textbf{M}}{\begin{bmatrix}4\lambda & 1 &-2 \\-2&(6\lambda-2) &-4\\-2 &+4&(8-2\lambda)\end{bmatrix}\begin{bmatrix}0 \\0\\0\end{bmatrix}}$$ I'm going to stop it here, partly so that you have fun math to finish up, partly because my break is over.

I hope this helps.

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    Hi Mark, Thanks for the detailed answer, I don't understand how did you get to this equation: $\begin{bmatrix}x_{critical}\\y_{critical}\\z{critical}\end{bmatrix}=\frac{1}{det\textbf{M}}{\begin{bmatrix}4\lambda & 1 &-2 \\-2&(6\lambda-2) &-4\\-2 &+4&(8-2\lambda)\end{bmatrix}\begin{bmatrix}0 \\0\\0\end{bmatrix}}$. If you can explain that, that would be great.. Also, would it be correct to find for which $\lambda$'s Det(M)=0, and then substitute that lambda in the equations and find the suitable x,y,z values? – 2017-02-01