2

So there is a rectangular box that has a volume of $8 m^3$. The top and bottom of the box is made with some material that has a cost of $8$ dollars per square meter. The sides are made with another material that costs $1$ dollar per square meter. How can I find the dimensions of the box that would minimize the total cost?

My thoughts:

I first started off with the equation, $V=lwh$. I think that we need to find an equation for cost in which we substitute $h$ into and then solve for two variables, then use the first derivative. I can get the concept but can't get how to work on the mechanics.

aña
  • 23
  • Can you write out a formula for the surface area? Then modify it to take the differing costs into account? – lulu Jul 14 '15 at 00:31
  • @ana: Consider that the box has 6 faces and there are 3 pairs of equal sides. – Gary. Jul 14 '15 at 00:34
  • @Gary Can you show what you mean? – aña Jul 14 '15 at 00:35
  • For the cost equation, consider what you are given: material is based on area. The top and bottom of the box are $lw$ square metres each, so the total cost for those is $16lw$. Then two sides are $lh$ square metres each, and the other two are $wh$ square metres each. Consider how to find total cost from this. – Terra Hyde Jul 14 '15 at 00:35
  • @aña: Assume you tear your box carefully, and open it up : notice the north ace and south face are equal to each other, and same goes for the left- and right - faces, and the two other faces. – Gary. Jul 14 '15 at 00:36
  • Hint: Lagrange Multipliers. – Archaick Jul 14 '15 at 00:36
  • @aña For calculations: Are you familiar with the $\texttt{Lagrange multipliers method}$ ? – callculus42 Jul 14 '15 at 00:37
  • @calculus I have heard of it, but I'm not that familiar. Can you perhaps show me? – aña Jul 14 '15 at 00:41
  • @aña You do not necessarily need the Lagrange multiplier method. You also can solve $abc=8$ for a and insert the term for a in the objective function. Then calculate the two derivatives w.r.t b,c and set the derivatives equal to zero. Then solve b and c. – callculus42 Jul 14 '15 at 00:50

2 Answers2

2

enter image description here

Let $a$ and $b$ denote the horizontal sides of the rectangular box and $c$ its height. One seeks to solve the following problem: \begin{align*} \min_{(a,b,c)\in\mathbb R_+^3}&\left\{8(2ab)+1(2ac+2bc)\right\}\\\text{s.t.}\quad&\,abc=8. \end{align*} Substitute for $c$ using the constraint $c=8/(ab)$ and plug it into the minimand to get, after some rearrangement: $$16 ab+\frac{16}{b}+\frac{16}{a}.$$ I leave it to you to check that this function is minimized at $a=b=1$ (note that it is now an unconstrained minimization problem). Hence, the optimal solution is $$(a^{\star},b^{\star},c^{\star})=(1,1,8).$$

triple_sec
  • 23,377
  • So how do I find the dimensions? – aña Jul 14 '15 at 00:49
  • The way it was set up, $a$ and $b$ are the length and width with $c$ being the height. – Terra Hyde Jul 14 '15 at 00:55
  • @aña I added a figure. – triple_sec Jul 14 '15 at 00:57
  • What does s.t. mean? – aña Jul 14 '15 at 00:57
  • @aña “Subject to” or “such that.” It's a standard abbreviation in constrained optimization problems. – triple_sec Jul 14 '15 at 00:58
  • So how do I see if it's minimized? – aña Jul 14 '15 at 00:58
  • @aña A necessary condition is that \begin{align} \frac{\partial}{\partial a}\left(16 ab+\frac{16}{b}+\frac{16}{a}\right)=&,0,\\frac{\partial}{\partial b}\left(16 ab+\frac{16}{b}+\frac{16}{a}\right)=&,0. \end{align} Sufficiency is trickier, as the function $(a,b)\mapsto 16ab+16/b+16/a$ is not convex on the positive orthant of the plane. Honestly, I didn't check sufficiency rigorously, but I just graphed the function. Based on the graph, it is apparent that it is minimized at a unique point. – triple_sec Jul 14 '15 at 01:02
  • Is there another way to show – aña Jul 14 '15 at 01:05
  • @aña I tried coming up with a two-step optimization algorithm involving only univariate calculus, but I'm afraid it's no less complicated than the one I had already posted. – triple_sec Jul 14 '15 at 02:28
0

Given a box of dimensions $l,w,h$ respectively, the surface area of the box is the sum of the surface areas of the top and bottom, given by $lw+lw =2lw$ plus the surface areas of the other 4 faces: $2lh+2wh$ , for a total of $$2lw+2lh+2wh$$. This means that the total cost of he material is $8$(# of meters in top and bottom)+ 1(material on the sides)=$$8(2lw)+1(lh+wh)= 16lw+lh+wh$$ But we must do this while satisfying the condition $lhw=8$ You can do this by Lagrange multipliers, as others pointed out, or you can substitute $lhw=8$ into the original problem and turn the problem into a problem of 2 variables, which I will do here, since others have done examples with Lagrange multipliers: $$16lw+lh+wh $$ becomes then $$ f(h,w)=\frac {128}{h}+\frac{8}{w}+hw= \frac{128w+ 8h +h^2w^2}{hw}$$

Now you have a function of 2 variables and you can use the techniques to find its maxmum and minimum values:

We start considering the partials $f_h, f_w$ and we set them equalto $0$, to find the critical points and we consider the discriminant $D$, with $$D=f_{xx}f_{yy}- f^2_{xy} $$.

And we consider the possibilities:

1)$D>0, f_{xx} <0$ 2)$D>0, f_{xx}>0 $

(There are other possibilities, but let's just consider these for now ).

In the case of $1$ we get a relative maximum, in the case of $2$ we get a relative minimum.

Can you see how to do it?

EDIT: The critical points are $y= 8/x^2 , x= 128/y^2$ , so that

Gary.
  • 2,432