Maximum length perimeter of a box whose diagonal is 10 unit long. Suppose the diagonal of a three-dimentional box has length $10$, what is its maximum perimeter length?
Here is my solution:
Let the three edges of the box adjacent to a vertex be labelled $a,b,c$. Then by Cauchy-Schwarz inequality, we have the size of the perimeters of the box is $$4 \cdot \langle(1,1,1),(a,b,c)\rangle \le 4 \sqrt{a^2+b^2+c^2}\sqrt{1^2+1^2+1^2}=4 \cdot 10 \cdot \sqrt3.$$ Hence we know that the size of the perimeter of the box is bounded by $40\sqrt3$. Then I still need to find an actual box with perimeter that value and with the size of its diagonal equals to $10$.

You want to maximise $a+b+c$ subject to the constraint $a^2+b^2+c^2=100$. This constraint is the surface of a sphere with radius $10$, so it is fairly clear from geometrical considerations that the maximum occurs when $a=b=c=\sqrt{\frac{100}{3}}$.
To show this formally, we can use Lagrange multipliers: let $f(a,b,c)=a+b+c$ and $g(a,b,c)=a^2+b^2+c^2-100$. Then we have to find $\lambda,a,b,c$ such that the Lagrange function
$$\left(\frac{\partial f}{\partial a},\frac{\partial f}{\partial b},\frac{\partial f}{\partial c}\right)-\lambda\left(\frac{\partial g}{\partial a},\frac{\partial g}{\partial b},\frac{\partial g}{\partial c}\right)$$
is zero. This gives
$$(1,1,1)-\lambda(2a,2b,2c)=0$$
So $$a=b=c=\frac{1}{2\lambda}$$
The constraint $g(a,b,c)=0$ gives us $\lambda=\pm\sqrt{\frac{3}{400}}$, and we get the solutions
$$(a,b,c)=\left(\sqrt{\frac{100}{3}},\sqrt{\frac{100}{3}},\sqrt{\frac{100}{3}}\right)$$
and
$$(a,b,c)=\left(-\sqrt{\frac{100}{3}},-\sqrt{\frac{100}{3}},-\sqrt{\frac{100}{3}}\right)$$
corresponding to the maximum and minimum values of $a+b+c$.
