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A tent with 2 rectangle shaped sides (no floor) and 2 isosceles triangles shaped gables with the volume $V$ is to be constructed. Determine the height so that the minimum amount of cloth is needed.

The tent is a prism with isosceles triangle bases. Let the height of the triangle (i.e. the height of the tent) be denoted $x$, the base $2y$ and the length of the tent $L$. Then the volume is $$V(x,y,L)=xyL$$ and the area of the tent will be $$A(x,y,L) = 2(xy+L\sqrt{x^2+y^2}).$$

Since A is a continuous function on a compact set (or can this actually be said, since V is not a boundary but a function of the variables?), there will be a minimum and maximum value. These are found when $grad \, V \, || \, grad \, A$. Since $$ grad \, V = (yL, xL, xy)$$ and $$grad \, A =2(y + \frac{xL}{\sqrt{x^2+y^2}}, x + \frac{yL}{\sqrt{x^2+y^2}}, \sqrt{x^2+y^2})$$ we must find $\lambda$ such that \begin{cases} y+\frac{xL}{\sqrt{x^2+y^2}} = \lambda yL \\ x+ \frac{yL}{\sqrt{x^2+y^2}} = \lambda xL \\ \sqrt{x^2+y^2} = \lambda xy \end{cases}

I have no idea how to solve this or if this even would be the correct approach. Any help is appreciated.

Lozansky
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  • Hint: It is $$\begin{cases} \color{red}{2}y+\color{red}{2}\frac{xL}{\sqrt{x^2+y^2}} = \lambda yL \ \color{red}{2}x+ \color{red}{2}\frac{yL}{\sqrt{x^2+y^2}} = \lambda xL \ \color{red}{2} \sqrt{x^2+y^2} = \lambda xy \end{cases}$$ You have forgotten the factor 2 of grad A. – callculus42 May 27 '15 at 17:06
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    Doesn't matter since it is multiplied by the same constant in all equations we can include it in $\lambda$. – Lozansky May 27 '15 at 17:50
  • It does matter. From these two equation systems you get different values for $\lambda$, $\lambda$ has a defined meaning. You can ignore a factor ($\neq 0$), if it is on both sides of an equation. – callculus42 May 27 '15 at 18:01
  • No it doesn't matter because two parallel vectors will still be parallel if you multiply either with a constant. In the end, this will not give a different solution because the 2's will eventually cancel each other out yielding the exact same answer. – Lozansky May 27 '15 at 18:17
  • All what I´m saying, that the values for Lambda will be different. The 2´s are cancelling out, if you divide the first equation by second equation. But Lambda will be cancelled out, too. At the end your value for Lambda will be twice as big, because you include the 2-as you said. – callculus42 May 27 '15 at 18:34
  • Why would you divide the first equation with the second? Yes $\lambda$ will be changed by a factor 2 because each equation is multiplied by a factor 2. Thus we will get the same values for $x,y,L$ which is what we are after. – Lozansky May 27 '15 at 19:12
  • It was only an example. I would put 2y (first equation) and 2x (second equation) on the RHS. Then dividing the first equation by the second equation, which lead to $\frac{x}{y}=\frac{y(\lambda L-2)}{x(\lambda L-2)}$. Cancelling out $\lambda L-2$ gives $x^2=y^2$ Because of $x,y>0$ it follows $x=y$ – callculus42 May 27 '15 at 21:15

2 Answers2

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You can continue with your approach. Adding the first two equations together: $$(x+y)(1+\frac{L}{\sqrt{x^2+y^2}}-\lambda L)=0$$

This gives you either $x=-y$ or $\lambda=\frac{1}{L}+\frac{1}{\sqrt{x^2+y^2}}$. Since $x=-y$ is not possible, we will continue with the other one. Plugging that into equation 1 or 2, you will get $x=y$.

Now if you replace all $x$ with $y$, you can get a relationship between $x$ and $L$. Remember that $V=xyL$ and $V$ is a constant. With that you can find $x,y$ and $L$.

I don't think your argument with bounded region holds for the minimum because the region is not bounded. It is kind of a hyperbaloid. I think you can either draw a rough picture to see there is a minimum, or use second derivative test on the $2D$ function $$A(x,y,L) = 2(xy+L\sqrt{x^2+y^2})=2(xy+\frac{\sqrt{x^2+y^2}}{xy})$$ to test the minimum.

KittyL
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  • I got that $x=y$ by replacing $L$ with $\frac{V}{xy}$ before but I made algebraic mistake when trying to express $L$ solely in $x$ and $y$. Is it a hyperboloid because $V$ is a function (albeit a constant one) so that the boundary can be rewritten to the form $x^2/a^2+y^2/b^2-z^2/c^2=1$? Or how did you realize the region is not bounded? – Lozansky May 27 '15 at 16:48
  • @Lozansky: Write it this way: $L=\frac{V}{xy}$. It is analogous to the $2D$ case $y=\frac{1}{x}$. To see that it is not bounded, you can let one of $x,y,L$ approach $0$, and another has to approach infinity to get the $V$. – KittyL May 27 '15 at 16:53
  • I didn't realize this because $L,x,y>0$ but this does make sense. – Lozansky May 27 '15 at 17:54
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Because volume V is fixed or constrained,

L = V/xy

substitute above eq into Area eq,

A(x,y) = 2(xy+V/xy*(x^2+y^2)^1/2)

this is minimization problem wrt x and y for area with fixed volume constraint. x, y within (0, inf)

because there are 2 indep variable, not only are you solving for tent height, but also tent base.

i made surface plot for Volume=5, to graphically see where/if the min is

enter image description here

based on plot, there do not seem to single min point, but may have multiple min point.

you can analytically for min point taking grad(A) wrt x and y and then solve for when grad(A) = 0 vector. you may find no single min point exist

classics
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  • Isn´t only x the independent variable ? – callculus42 May 27 '15 at 17:45
  • if there are no constraint, then there would 3 variable determining tent shape (x, y, L). but because volume is constrained, L can be expressed as function of x and y. thus reducing number of indep variables to x and y. Because there are no additional constraint, x and y are free change. thus this is minimization of 2 variables instead of one. – classics May 27 '15 at 18:14