extrema under constraints - lagrangian multipliers

Question

Find all global extrema of $f(x,y,z)=x^3+y^3+z^3$ under the constraints

• a) $x^2+y^2+z^2=1$
• b) $2x^2+y^2=1$

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Regarding a), I've tried to use the lagrangian multipliers, that is $(3x^2-\lambda 2x,3y^2-2\lambda y-3z^2-2\lambda z,-x^2-y^2-z^2+1)=(0,0,0,0)$ and solving this gives me $x=0$ or $x=\frac{2}{3}\lambda$. If I try to substitute the latter in the other equations, I only get a solution of the system that is dependent on $\lambda$, but no actual solution for $\lambda$ itself. The same issue arises when I try to solve b).

Answer 1

That is not how the method of Lagrange multipliers should be used whan there is mor than one restriction. You should solve the system$$\left\{\begin{array}{l}\frac{\partial f}{\partial x}(x,y,z)=\lambda\frac{\partial g}{\partial x}(x,y,z)+\mu\frac{\partial h}{\partial x}(x,y,z)\\\frac{\partial f}{\partial y}(x,y,z)=\lambda\frac{\partial g}{\partial y}(x,y,z)+\mu\frac{\partial h}{\partial y}(x,y,z)\\\frac{\partial f}{\partial z}(x,y,z)=\lambda\frac{\partial g}{\partial z}(x,y,z)+\mu\frac{\partial h}{\partial z}(x,y,z)\\g(x,y,z)=1\\h(x,y,z)=1.\end{array}\right.$$You will get lots of solutions:

• $(x,y,z)=(0,\pm1,0)$;
• $(x,y,z)=\pm\left(\frac1{\sqrt2},0,\frac1{\sqrt2}\right)$;
• $(x,y,z)=\pm\left(\frac1{\sqrt2},0,-\frac1{\sqrt2}\right)$;
• $(x,y,z)=\pm\left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right)$.