The problem (from Calculus for Business, Economics, Life Sciences and Social Sciences 12e):
I found this and that, but they use Lagrange/KKT.
What I tried:
Girth $= 2w + 2h$
Maximise $V(l,w,h) = lwh$ s.t.
$$l + 2w + 2h \le 120 \tag{*}$$
$$\to l \le 120 - (2w + 2h)$$
$$\to V=lwh \le (wh)(120 - (2w + 2h))$$
$$\to V \le 120wh - 2w^2h - 2wh^2$$
Then I don't know. If $(*)$ was an equality instead, what I would do is:
Maximise $V(l,w,h) = lwh = U(w,h) := (120 - (2w + 2h))(wh)$
This is a strange problem. Why are we not allowed to use optimization methods to solve optimization problem? But anyway ...
First, use the geometric result that the area of a rectangle is maximized subject to perimeter constraint is a square (this itself need a proof using calculus), but look like we are allowed to use calculus so we will take this for granted.
So the problem becomes maximize $V = a^2 l$ subject to $4a + l \le 120$. Again, geometrical insight helps. There is no reason to believe the equality not holding will give a maximum because increasing the length will always help (unlike in profit maximisation problems where increasing revenue doesn't necessarily increase profit).
So we set the equality and eliminate l to get maximize $V = a^2 (120 - 4a)$, this is just plain calculus and we get
$$V' = 240a - 12a^2 = 0$$ $$a = 20$$ $$l = 120 - 4(20) = 40$$
So that's the answer - without using Lagrange multiplier or KKT, but we used a few geometrical 'insight' without proof.