0

The problem (from Calculus for Business, Economics, Life Sciences and Social Sciences 12e):

enter image description here

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 $V(l,w,h) = lwh = U(w,h) := (120 - (2w + 2h))(wh)$

Community
  • 1
BCLC
  • 13,459

1 Answers1

1

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.

BCLC
  • 13,459
Andrew Au
  • 1,127
  • It is from a basic calculus book in the section immediately preceding the section on lagrange multipliers. Anyway thanks Andrew. Going to read now – BCLC Jan 21 '16 at 08:05
  • Brilliant insight on the rectangle thing. What exactly is meant by 'Again, geometrical insight helps. There is no reason to believe the equality does not hold, adding the length will always help, right?' ? You mean choosing $a$ and $l$ s.t. $4a + l =120$ will indeed maximise the volume? This is because...$a$,$l$ and $V$ are positive quantities? – BCLC Jan 21 '16 at 08:07
  • 1
    Yes, there is no reason to choose $ a $ or $ l $ in a way such that $ 4a + l $ is strictly less than 120, for you can always increase $ l $ to further increase the volume. Therefore, in the sequel, I assumed $ 4a + l = 120 $. – Andrew Au Jan 22 '16 at 06:03
  • Oh that's strange. That's what I thought of later on without checking. I thought that that was what I commented earlier. Anyway thanks ^-^ – BCLC Jan 22 '16 at 06:51
  • Actually need we use the fact about $a=w=h$? I think it's something we'll encounter anyway if we simply replace inequality with equality in constraint – BCLC Jan 22 '16 at 07:04