I have a problem
"What is the maximum possible volume of a rectangular box that can be inscribed in the unit sphere?"
The thing to maximize is $8xyz$ (volume of a sphere with coords $(x,y,z)$) and the restraints are $x^2 + y^2 + z^2 = 1$
The equations I got are
$8yz = \lambda * 2x$
$8xz = \lambda * 2y$
$8xy = \lambda * 2z$
Which can be solved for their respective variables and plugged into the restraint to get
$(4yz)^2 + (4xz)^2 + (4xy)^2 = \lambda ^2$
Which (as far as I can tell) isn't easily solvable.
Did I miss a step or confuse the process? It would be super helpful if someone could walk through the problem
Just a picture, not an answer. 
Best Answer
So let me give you a general case.
Let $(x,y,z)$ be the corner of the box in the positive octant, with sides parallel to the axes. Then the volume of the box is $8xyz$. The sphere of radius $a$ is given by $x^2+y^2+z^2=a^2$.
So to answer the question we want to maximise the function:
$\displaystyle f(x,y,z)=8xyz$
with the constraint:
$\displaystyle g(x,y,z)=0$ where $\displaystyle$ $g(x,y,z)=x^2+y^2+z^2-a^2.$
Using the method of Lagrange multipliers we know:
$\displaystyle\nabla f(x,y,z)=\lambda\nabla g(x,y,z)$ at the maximum.
Calculating the gradients:
$\displaystyle\nabla f(x,y,z)=(8yz,8xz,8xy)$, and
$\displaystyle\nabla g(x,y,z)=(2x,2y,2z)$
So we have:
$\displaystyle 8yz=2\lambda x,$ $\displaystyle 8xz=2\lambda y,$ $\displaystyle 8xy=2\lambda z$ Dividing 1. by 2. we obtain:
$\displaystyle\frac{y}{x}=\frac{x}{y} \quad \Longrightarrow \quad x^2=y^2$
Similarly dividing 2. by 3. we get $y^2=z^2$, so that means $x^2=y^2=z^2$
Returning to our constraint equation, $g(x,y,z)=0$, we obtain:
$\displaystyle 3x^2=a^2 \Longrightarrow x=\frac{a}{\sqrt{3}} $
Which means $x=y=z=\frac{a}{\sqrt{3}}$
So returning to the volume, the maximal volume is:
$\displaystyle f\left(\frac{a}{\sqrt{3}},\frac{a}{\sqrt{3}},\frac{a}{\sqrt{3}}\right)=\frac{8a^3}{3\sqrt{3}}$
where $a$ is the radius of the sphere.
Now find the volume in a unit sphere.