Find the best dimensions of a tent with horizontal ridge pole. Assume the two ends closed by isosceles triangles. There is no floor.

The total surface area of the tent is \begin{align*} A & = 2 \times \text{Area of the rectangular sides} + 2 \times \text{Area of the isosceles triangles at the ends}. \end{align*} Let $x$ and $y$ be the sides of the rectangular canvas and $\theta$ be the angle made by the isosceles triangle. Then the area of the tent is \begin{align*} A & = 2xy+2 \times \frac{x^{2}}{2}\sin \theta = 2xy+x^{2}\sin \theta. \end{align*} For the minimum of $A$, we have the necessary conditions $\displaystyle \frac{\partial A}{\partial x}=0$, $\displaystyle \frac{\partial A}{\partial y}=0$ and $\displaystyle \frac{\partial A}{\partial \theta}=0$. Therefore, differentiating $f$ with respect to $x$, $y$ and $\theta$, we have \begin{align*} \frac{\partial A}{\partial x} & = 2y+2x \sin \theta, \quad \frac{\partial A}{\partial y} = 2y \quad \text{ and } \quad \frac{\partial A}{\partial \theta} = 2x\cos \theta. \end{align*}

Is the formulation correct? If not please provide the correct formulation. Thank you.

  • $\begingroup$ What do you mean by a "tent with no floor" ? That the two sides of the roof reach the ground ? $\endgroup$
    – Jean Marie
    Commented Feb 27 at 16:54
  • $\begingroup$ No floor means - no canvas is used for the floor. $\endgroup$ Commented Feb 27 at 17:17
  • $\begingroup$ The first sentence "find the best dimensions..." doesn't give all the information ! Do you want to minimize the total canvas area for a given volume ? But here you don't mention the volume ?.... $\endgroup$
    – Jean Marie
    Commented Feb 27 at 18:10
  • $\begingroup$ Please say if the question of your homework mentions the volume of the tent. This should be the case. Because if you want to minimize the area, take it at once to be equal to zero... $\endgroup$
    – Jean Marie
    Commented Feb 28 at 17:39
  • $\begingroup$ I have got the problem from Advanced Calculus, Widder. As you said it should be with minimum surface area for the given volume. $\endgroup$ Commented Feb 29 at 5:38


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