Jtdylktuy

Suppose the diagonal of a 3 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-schwarts inequality, we have the size of the perimeters of the box is $4* 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*10*sqrt(3) $. Hence we know that the size of the perimeter of the box is bounded by $40*sqrt(3)$. 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 $a=b=c=sqrt{frac{10}{3}}$.

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

$$left(frac{partial f}{partial a},frac{partial f}{partial b},frac{partial f}{partial c}right)-lambdaleft(frac{partial g}{partial a},frac{partial g}{partial b},frac{partial g}{partial c}right)$$
is zero. This gives
$$(1,1,1)-lambda(2a,2b,2c)=0$$
So $$a=b=c=frac{1}{2lambda}$$
The constraint $g(a,b,c)=0$ gives us $lambda=pmsqrt{frac{3}{40}}$, and we get the solutions
$$(a,b,c)=left(sqrt{frac{10}{3}},sqrt{frac{10}{3}},sqrt{frac{10}{3}}right)$$
and
$$(a,b,c)=left(-sqrt{frac{10}{3}},-sqrt{frac{10}{3}},-sqrt{frac{10}{3}}right)$$

corresponding to the maximum and minimum values of $a+b+c$.

hint

$$L^2+l^2=D^2=100$$

the perimeter is

$$P=2(L+l)$$

Let maximise $P^2$.

$$P^2=400+4Lsqrt{100-L^2}$$

Solve for $L$ the equation
$$P'(L)=0$$

which gives

$$sqrt{100-L^2}-frac{2L^2}{2sqrt{100-L^2}}=0$$

You should find $$L=5sqrt{2}=l$$

It is a square of maximal perimeter $$P_{max}=20sqrt{2}$$