High School Calculus: Find cut out length x which gives Maximum volume of can.

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This is a pretty long question so I will try to shorten it as much as possible.

Part A:

Given: a square piece of tin plate with side length $l$, cut out corners with shape of square of side length $x$. Fold the resulting shape into an open-ended cuboid. Determine $x$ for maximum volume (deduce conjecture)

Next, prove, using the principles of calculus (and the quadratic formula),the conjecture.

Pictues of the problem that I drew: https://i.stack.imgur.com/YvXk5.png

What I did:

$V(x)=Length×Width×Height$ (Equation for cuboid)

$V(x)=x(l-2X)(l-2x)$ (substitute known values)

$=x[(l-2X)]^2$ $=x[(l-2X)]^2$ $=x(l^2-4lx+4x^2)$ $=l^2 x-4lx^2+4x^3$

$V'(x)=l^2-8lx+12x^2$ (differentiate)

$0=l^2-8lx+12x^2$ ($V'(x)=0$ to find out maximum/minimum)

$0=(2x-l)(6x-l)$ (factorise)

$2x-l=0$

$2x=l$

$x=l/2$ (first answer is not correct, if $x=1/2$ then cuboid cant even be made)

$6x-l=0$

$6x=l$

$x=l/6$ (correct answer)

I am unable to do the next part, Im not sure if i am overthinking it or not. But I cant seem to figure out how do i prove this conjecture? Haven't i already proved it?

Part B

https://i.stack.imgur.com/jxYWf.png

It ask the same thing as above $x$ when volume is at maximum, however, as you can see from this picture. This time it is a rectangular plate with side length $r$ and $s$.

What I did:

$V(x)=x(r-2x)(s-2x)$ (substitute values into equation)

$=x(rs-2rx-2sx+4x^2)$

$=rsx-2rx^2-2sx^2+4x^3$

$V'(x)=rs-4rx-4sx+12x^2$ (differentiate)

$=rs-4x(r+s)+12x^2$ $=12x^2-4x(r+s)+rs$

$0=12x^2-4x(r+s)+rs$ (set $V'(x)$ as $0$ to find out maximum/minimum)

$x=(--4(r+s)±√([(-4(r+s))]^2-4\times12rs))/(2\times12)$ (use quadratic formula as it is impossible to factorize)

$x=(4(r+s)±√([(-4(r+s))]^2-48rs))/24$

$=(4(r+s)±√([16(r+s)]^2-48rs))/24$

$=(4(r+s)±√([16[(r+s)]^2-3rs]))/24$

$=(4(r+s)±4√([(r+s)^2-3rs]))/24$

$=((r+s)±√([(r+s)]^2-3rs))/6$

$=((r+s)±√(r^2+2rs+s^2-3rs))/6$

$=(r+s±√(r^2-rs+s^2 ))/6$

And from what I think (only when the $+√(r^2-rs+s^2$ )gives the maximum volume)

But again, I am suppose to prove the conjecture that I just deduced.

Just How?

Any advice are welcome.