Find absolute maxima and minima of f(x,y,z)=xyz subject to g(x,y,z)=x^2+y^2+z^2-12 and h(x,y,z)=x+y+z-4

Question

I had no problem getting these equations:

$yz=2\lambda x + \mu$
$xz=2\lambda y + \mu$
$xy=2\lambda z + \mu$
$x^2+y^2+z^2=12$
$x+y+z=4$

The part that I can not figure for the life of me is how to solve these for x,y,z. Any help?

Answer 1

Here’s a relatively straightforward solution that doesn’t use Lagrange multipliers.

Notice that because of the symmetry of $f$ and the two constraints, if $f$ has an extremum at the point $(a,b,c)$, then it also has an extremum of the same type at all of the points with coordinates that are permutations of $(a,b,c)$.

The two constraints taken together specify the intersection of a sphere around the origin with radius $\sqrt{12}$ and the plane $x+y+z=4$, i.e., a circle in this plane centered at $P_0=\small\left(\frac43,\frac43,\frac43\right)$ with radius $R=\sqrt{12-||P_0||^2}=\small\sqrt\frac{20}3$. The position of this circle and symmetry of $f$ suggest that the critical points will come in equally-spaced sets of three, and that one set of extrema will be at the points of the circle that are closest to the “corners,” with another set of extrema midway between them:

We proceed by finding a parametrization $\phi:\mathbb R\to\mathbb R^3$ of this circle, thereby reducing the problem to finding the extrema of $f\circ\phi$. One way to derive $\phi$ is to take the same-sized circle centered on the $z$-axis in the plane $z=||P_0||=\small\frac4{\sqrt3}$ and rotate it into position, i.e., rotate the $z$-axis so that it coincides with $P_0$. One such rotation is $$ M=\pmatrix{ \frac16(\sqrt3+3) & \frac16(\sqrt3-3) & \frac1{\sqrt3} \\ \frac16(\sqrt3-3) & \frac16(\sqrt3+3) & \frac1{\sqrt3} \\ -\frac1{\sqrt3} & -\frac1{\sqrt3} & \frac1{\sqrt3}}, $$ which gives $$ \phi:t\mapsto M\pmatrix{R\cos t \\ R\sin t \\ ||P_0||}=\small \frac16\pmatrix{ (\sqrt3+3)\;R\cos t + (\sqrt3-3)\;R\sin t + 2\sqrt3\;||P_0|| \\ (\sqrt3-3)\;R\cos t + (\sqrt3+3)\;R\sin t + 2\sqrt3\;||P_0|| \\ -2\sqrt3\;(R\cos t + R\sin t - ||P_0||) }. $$ Some tedious algebraic manipulation yields $$f(\phi(t))=\frac1{6\sqrt3}(R^3\cos{3t}+R^3\sin{3t}+2\;||P_0||^3-3R^2\;||P_0||).$$ This has maxima at $\frac13(-\frac\pi4+2\pi k)$ and minima at $\frac13(\frac34\pi+2\pi k)$, confirming the threefold symmetry guessed at above. Plugging these values into $\phi$ gives maxima at the points $$\small \left(\frac23(2+\sqrt{10}),\frac13(4-\sqrt{10}),\frac13(4-\sqrt{10})\right) \\ \small \left(\frac13(4-\sqrt{10}),\frac23(2+\sqrt{10}),\frac13(4-\sqrt{10})\right) \\ \small \left(\frac13(4-\sqrt{10}),\frac13(4-\sqrt{10}),\frac23(2+\sqrt{10})\right) $$ and minima at $$\begin{align} &\small \left(\frac23(2-\sqrt{10}),\frac13(4+\sqrt{10}),\frac13(4+\sqrt{10})\right)\\ &\small \left(\frac13(4+\sqrt{10}),\frac23(2-\sqrt{10}),\frac13(4+\sqrt{10})\right) \\ &\small \left(\frac13(4+\sqrt{10}),\frac13(4+\sqrt{10}),\frac23(2-\sqrt{10})\right), \end{align}$$ which exhibit the expected permutation symmetries.

Finally, the maximum value of $f$ with the given constraints is $-\frac4{27}(14-5\sqrt{10})\approx 0.268$ and the mimimum is $-\frac4{27}(14+5\sqrt{10})\approx -4.417$.