A carpenter wants to manufacture a box with a volume of $1 \text{m}^3$. The box has sides parallel to the coordinate plane, and the side lengths are given by $x$, $y$, and $z$, where $0 \leq x \leq 10$, $0 \leq y \leq 10$, and $0 \leq z \leq 10$. The front and top sides of the box will be made of an expensive type of wood that costs $900$ per $\text{m}^2$, while the bottom and other sides will be made of a cheaper type of wood that costs $300$ per $\text{m}^2$. The carpenter wants to choose the side lengths to minimize the cost.

a) Formulate the carpenter's problem mathematically as a minimization problem with constraints.

b) Use the Lagrange multiplier method to solve the minimization problem in part (a) of the question. No points will be awarded for using other methods.

A) :

$$f(x,y,z) = 1200xz+1200xy+600yz$$ $$g(x,y,z) = xyz = 1$$ My expression of the the function was correct but when it comes to the constraints there is something wrong.

B) Here I used my constraint and function and used Lagrange multipliers to get:

$$x= 2^{-\frac23}$$ $$y= 2^{\frac13}$$ $$z= 2^{\frac13}$$ $$f(x,y,z) = 1800 \times 2^{\frac23} $$

So far I'm correct and I thought it was the answer but they continued in the answers They said:

Here we have to check edge points, i.e. $x = 10, y = 10 or z = 10. If x = 10$, then we get minimization problems.

After this they did a computation that took over one page to complete but I couldn't understand anything because I'm totally confused. Why aren't we done where I was "done"? What is my constraint to begin with? What is the point of $0 \leq x \leq 10$, $0 \leq y \leq 10$, and $0 \leq z \leq 10$?


1 Answer 1


The point of the constraints $0 \leq x \leq 10$ etc. is to make the problem not silly (read: maybe make it realistic?), in two ways.

First, the lower bound of $0$ both makes us think about physical quantities (what does it mean for a box to have negative width?) and prevents us from just taking e.g. $x \to -\infty$, which definitely minimises the cost, but in a way that's totally meaningless.

Second, the upper bound of $10$ is about making the problem physically meaningful even after we decide negative quantities are silly. Without it, we could for example let $x$ and $y$ be really, really close to $0$ and then let $z$ blow up (so as to still satisfy $x y z = 1$). This kind of thing may or may not be an issue when optimising (in this case it is not; the cost grows big as you do this), but either way it's somehow unrealistic to build an arbitrarily tall but skinny box (for instance). Think of this as perhaps the carpenter doesn't have too large a sheet of the material for each side (maybe it comes in $10$ by $10$ metre sheets).

With all of this said, the reason they go on to check the endpoints ($x = 10$ etc.) isn't that your answer is incorrect (it's not), it's that without checking the endpoints we can't know if we've found a local or a global extremum. Lagrange multipliers find a particular kind of local extrema, so it could be that whilst your calculations are correct, a point with lower cost exists somewhere undetectable by Lagrange multipliers.

This is similar to how, for example, if you want to find the global minimum of, say, $f(x) = -x^3 + x^2 + x$ on $[{-2},2]$, you compute the derivative, find where it's zero ($x = -1/3$ and $x = 1$), figure out that at $x = -1/3$ the function is concave up (or some other similar strategy) and think "then it must be the minimum I want..!" except of course that's not quite right; you've only found a local minimum, but it's possible the constraints $x \in [{-2},2]$ gives minima (and indeed they do in this made up example).

  • $\begingroup$ Firstly thanks for this explanation. Secondly there are still things I don't understand, I personally assumed $xyz = 1$ was my constraint. Is it not? In that case what is $xyz = 1$ in my computation then? Does it mean I have two constraints? On a second thought $xyz = 1$ doesn't feel like a constraint (never felt) but then I don't know how to do optimization of intervals. $\endgroup$ May 24, 2023 at 14:43
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    $\begingroup$ Both $xyz=1$ and the inequalities $0 \leq x \leq 10$ etc. are constraints. Certainly $xyz=1$ does constrain you: without it you could make $x$, $y$, and $z$ anything, with it you can't; likewise without $0 \leq x \leq 10$ you could make $x$ anything, with it you can't. $\endgroup$ May 24, 2023 at 14:50
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    $\begingroup$ The big picture point is that Lagrange multipliers is a tool for handling constrained optimisation problems subject to equality constraints (like $xyz \color{red}{=} 1$), not inequality constraints (like $0 \color{red}{\leq} x \color{red}{\leq} 10$), so your Langrange multiplier approach doesn't know about the inequalities at all. That's why we handle them separately at the end. $\endgroup$ May 24, 2023 at 14:53
  • $\begingroup$ Okay so we have two different constraints right? Then I really don't understand your analogy (derivative and the interval in the single variable example). I computed a constraint right but in your single variable example, what does this symbolize? $f'(x)=0$? But then what does $\nabla f(x,y,z) = (0,0,0)$ symbolize in this case? The question specifically says to use lagrange so I ignored this part but why is the partial derivative being 0 useless? $\endgroup$ May 24, 2023 at 15:41
  • $\begingroup$ The purpose of the analogy is to explain why we need to think about the $0 \leq x \leq 10$ constraints, that's all. The Lagrange multipliers are analogous to finding the critical points $f'(x) = 0$, but don't check the endpoints at all. $\endgroup$ May 24, 2023 at 15:44

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