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.