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spin [16.1K]
3 years ago
14

Use lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f(x,y = xyz; x^

2 2y^2 3z^2=96
Mathematics
1 answer:
Snezhnost [94]3 years ago
6 0
I'm assuming the constraint involves some plus signs that aren't appearing for some reason, so that you're finding the extrema subject to x^2+2y^2+3z^2=96.

Set f(x,y,z)=xyz and g(x,y,z)=x^2+2y^2+3z^2-96, so that the Lagrangian is

L(x,y,z,\lambda)=xyz+\lambda(x^2+2y^2+3z^2-96)

Take the partial derivatives and set them equal to zero.

\begin{cases}L_x=yz+2\lambda x=0\\L_y=xz+4\lambda y=0\\L_z=xy+6\lambda z=0\\L_\lambda=x^2+2y^2+3z^2-96=0\end{cases}

One way to find the possible critical points is to multiply the first three equations by the variable that is missing in the first term and dividing by 2. This gives

\begin{cases}\dfrac{xyz}2+\lambda x^2=0\\\\\dfrac{xyz}2+2\lambda y^2=0\\\\\dfrac{xyz}2+3\lambda z^2=0\\\\x^2+2y^2+3y^2=96\end{cases}

So by adding the first three equations together, you end up with

\dfrac32xyz+\lambda(x^2+2y^2+3z^2)=0

and the fourth equation allows you to write

\dfrac32xyz+96\lambda=0\implies \dfrac{xyz}2=-32\lambda

Now, substituting this into the first three equations in the most recent system yields

\begin{cases}-32\lambda+\lambda x^2=0\\-32\lambda+2\lambda y^2=0\\-32\lambda+3\lambda z^2=0\end{cases}\implies\begin{cases}x=\pm4\sqrt2\\y=\pm4\\z=\pm4\sqrt{\dfrac23}\end{cases}

So we found a grand total of 8 possible critical points. Evaluating f(x,y,z)=xyz at each of these points, you find that f(x,y,z) attains a maximum value of \dfrac{128}{\sqrt3} whenever exactly none or two of the critical points' coordinates are negative (four cases of this), and a minimum value of -\dfrac{128}{\sqrt3} whenever exactly one or all of the critical points' coordinates are negative.
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<u>Pre-Algebra</u>

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