Find the constrained minimum and maximum of the following $f$

Question

So, I'm given the linear function $f(x,y,z)=8x-y+z$ and I need to find the minimum and maximum over the set $\Sigma=\{(x,y,z)\in\mathbb{R}^3\mid x^4+y^4+z^4=1\}$. Since the gradient of $f$ is never $0$, we can directly move the the system of equations:

$x^4+y^4+z^4=1$
$8=4\lambda x^3$
$-1=4\lambda y^3$
$ 1 = 4\lambda z^3$

coming due to Lagrange multipliers. Resolving this system, up to mistakes, gave me: $x=\frac{8\cdot2^{\frac{1}{3}}}{\alpha}$, $y=-z=\frac{1}{4^{\frac{1}{3}}}\frac{1}{\alpha}$ where $\alpha =\big(2^{\frac{4}{3}}+\frac{2}{4^{\frac{4}{3}}}\big)^{\frac{1}{4}}$. Probably there are some mistakes around the resolution, but up to the systems everything should work. Even if I'm not mistaken, what is the nature of the point I found? It should be a minimum or a maximum? And from where do I find the other point giving the other extreme?

Answer 1

That system has two and only two solutions:$$\left(\frac{\sqrt[4]8}{\sqrt3},-\frac1{\sqrt[4]2\sqrt3},\frac1{\sqrt[4]2\sqrt3}\right)\left(\longleftrightarrow\lambda=\frac{\sqrt{27}}{2\sqrt[4]2}\right)$$and$$\left(-\frac{\sqrt[4]8}{\sqrt3},\frac1{\sqrt[4]2\sqrt3},-\frac1{\sqrt[4]2\sqrt3}\right)\left(\longleftrightarrow\lambda=-\frac{\sqrt{27}}{2\sqrt[4]2}\right).$$If you compute the value of $f$ at the first solution, you will get a number greater than $0$, and if you compute the value of $f$ at the second solution, you will get a number smaller than $0$. On the other hand, since $\Sigma$ is closed an bounded, $f|_\Sigma$ must have a maximum and a minimum. And the fact that every point of $\Sigma$ is a regular point of $(x,y,z)\mapsto x^4+y^2+z^4$ assures us the method of Lagrange multipliers will allow us to find all such points. So, the first solution is the point at which the maximum is attained, and the second solution is the point at which the minimum is attained.

Answer 2

Define $f$ by $f(x,y,z)=8x-y+z$ and $g$ by $g(x,y,z)=x^4+y^4+z^4-1$. We seek the maximum and minimum of $f(x,y,z)$ subject to $g(x,y,z)=0$. If an extremum is attained at a point $(x,y,z)$ then by the method of Lagrange multipliers there exists $\lambda \in \mathbb{R}$ such that $$0 = \nabla f(x,y,z) + \lambda \nabla g(x,y,z) \ \mathrm{and} \ g(x,y,z)=0 $$ We see that $$0 = \nabla f(x,y,z) + \lambda \nabla g(x,y,z) = (8+4\lambda x^3,-1+4\lambda y^3,1+4\lambda z^3)$$ gives the unique solution $(x,y,z) = (x_\lambda,y_\lambda,z_\lambda) =\lambda^{-1/3}(-2^{1/3},4^{-1/3},-4^{-1/3})$. We see that $$0=g(x_\lambda,y_\lambda,z_\lambda) = (2^{4/3}+2\cdot 4^{-4/3})\lambda^{-4/3}-1 \Leftrightarrow \lambda = \pm (2^{4/3}+2\cdot 4^{-4/3})^{3/2} = \pm 27\cdot 2^{-5/2}.$$ which gives the corresponding extremal points $(x_\pm,y_\pm,z_\pm)=\pm (2^{4/3}+2\cdot 4^{-4/3})^{-1/2}(-2^{1/3},4^{-1/3},-4^{-1/3}) = \pm(2^{1/6}/3)(-2,1,-1)$. The corresponding values of $f$ are $f(x_\pm,y_\pm,z_\pm)= \mp 8 \cdot 2^{1/3} \cdot (2^{4/3}+2\cdot 4^{-4/3})^{-1/2}=\mp 2^{25/6}/3$. Thus, $(x_+,y_+,z_+)$ is a local minimum point with corresponding value $-2^{25/6}/3$ and $(x_-,y_-,z_-)$ is a local maximum point with corresponding value $2^{25/6}/3$.