Using Lagrange multipliers, I can see that the maximum of $f(x, y, z)$ subject to the constraint is $\sqrt{3}$.
To see this, here is what I did:
Show the constraint in the form $g(x,y,z) = k$, which gives us $g(x, y, z) = x^4 + y^4 + z^4 - 1$.
To satisfy $\nabla{f(x, y, z)} = \lambda \nabla{g(x, y, z)}$, start by taking all the relevant partial derivatives:
- $\frac{\partial f}{\partial x} = 2x$
- $\frac{\partial f}{\partial y} = 2y$
- $\frac{\partial f}{\partial z} = 2z$
- $\frac{\partial g}{\partial x} = 4x^3$
- $\frac{\partial g}{\partial y} = 4y^3$
- $\frac{\partial g}{\partial z} = 4z^3$
Set up a system of equations, and solve.
- $2x = \lambda4x^3$
- $2y = \lambda4y^3$
- $2z = \lambda4z^3$
- $x^4 + y^4 + z^4 = 1$
Skipping ahead through the remaining arithmetic, this gives $x^2 = y^2 = z^2$ and also $x = y = z = \sqrt{\frac{1}{3}}$.
To see this in more detail, we can use $g(x, y, z)$. If $x^2 = y^2 = z^2$ as we stated, then we have:
- $(x^2)^2 + (y^2)^2 + (z^2)^2 = 1$
and therefore
- $(x^2)^2 + (x^2)^2 + (x^2)^2 = 1$, etc.).
Plug this into $f(x, y, z)$ to get $\sqrt{3}$, which is the maximum of the function subject to the constraint.
However, the answer key in my textbook says that the minimum is $1$. Can someone explain why the minimum is $1$, and how to understand that?
$x = 2\lambda x^3$
Either $x = 0$ or $\lambda =\frac {1}{2x}$.
You have investigated what happens if $\lambda = \frac 1{2x}$ but what about $x = 0?$