Finding the maximum and minimum values of a function in 3 variables subject to a given constraint using Lagrange multiplier

Question

Question: Use Lagrange Multiplier to determine the maximum values of $f(x,y,z) = x^2 + y^2 + z^2$ subject to constraint $xyz = 4$.

I do not know how to solve this, I got the expression $x^2 = y^2 = z^2$ then what should I do next?

Thank you.

Answer 1

With

$$ f = x^2+y^2+z^2 $$

forming the Lagrangian

$$ L(x,y,z,\lambda) = f+\lambda(x y z-4) $$

and determining the stationary points by solving

$$ 2x+\lambda y z = 0\\ 2y+\lambda x z = 0\\ 2z+\lambda x y = 0\\ x y x - 4 = 0 $$

we get at

$$ \left[ \begin{array}{cccc} x & y & z & f\\ -2^{2/3} & -2^{2/3} & 2^{2/3} & 6 \sqrt[3]{2} \\ -2^{2/3} & 2^{2/3} & -2^{2/3} & 6 \sqrt[3]{2} \\ 2^{2/3} & -2^{2/3} & -2^{2/3} & 6 \sqrt[3]{2} \\ 2^{2/3} & 2^{2/3} & 2^{2/3} & 6 \sqrt[3]{2} \\ \end{array} \right] $$

NOTE

The system to solve is equivalent to

$$ 2x^2+\lambda x y z = 0\\ 2y^2+\lambda x y z = 0\\ 2z^2+\lambda x y z = 0\\ x y x - 4 = 0 $$

Answer 2

Since you already got $x^2=y^2=z^2$, use the constraint $xyz=4$. Square the constraint, and you get $x^2y^2z^2=16$. Plugging in your result, you get $x^6=2^4$. Then $x^2=2^\frac 43=2\sqrt[3]2$. Now using your result again in $f(x,y,z) =x^2+y^2+z^2$, you get the maximum as $3\cdot 2\sqrt[3]2=6\sqrt[3]2$/

Answer 3

HINT: The trick is to define a function $g(x, y, z) = f(x, y, z) -\lambda(xyz - 4)$ which takes the same values as $f(x, y, z)$ as long as the constraint is satisfied; $\lambda$ is the Lagrance multiplier. Now you maximize $g(x, y, z)$ without having to worry about the constraint. You do that by taking the partial derivatives wrt. $x, y, z$ and setting them equal to 0 to find the extrema. That's three equation in four unknowns ($x, y, z, \lambda$), but the constraint equation comes to the rescue providing a fourth equation (alternatively, you can consider $g$ as a function of $x, y, z\ AND\ \lambda$ and in addition, take the partial derivative of $g$ wrt $\lambda$ and set that equal to 0; but you can easily see that that's the constraint equation). Now solve them for the four unknowns and substitute the values in $f(x,y,z)$ to find the extreme values of $f$. You then have to worry about which one(s) give you the maximum value, but that should be easy: there is a finite number of solutions, so you can enumerate them and figure out which one(s) give you the maximum value.