Consider the following question
A tent manufacturer wants to maximize the volume of a new design of tent, subject only to a constant weight (which is directly proportional to the amount of fabric used). The models considered have either equilateral-triangular or semi-circular vertical crosssection, with vertical planar ends in both cases and with floors of the same fabric. Which shape maximizes the volume for a given area $A$ of fabric? $\left[\right.$ Hint: $(2 \pi)^{-1 / 2} 3^{-3 / 4}(2+\pi)<1$.]
This is in the context of Lagrange multipliers.
The way I am interpreting this is that the one with the triangle is a cone whilst the other is a semi-sphere. However, this can not be right. This is because if that is the case then given the area of fabric the volume is immediately specified, whereas the question should be examining use of Lagrange multipliers. This suggests I am missing some variation.
Could someone help me understand the models?
Note: I am not asking for the solution of the question. Just for help in understanding what it is saying.
I will answer the question in the case of a third tent shape, a rectangular prism, where the tent cross-sections are all squares. Hopefully, this will illustrate what the problem is asking, while still allowing you to solve the entire exercise yourself.
A rectangular prism with square cross sections is defined by two parameters; the side length of the square, $s$, and the length of the tent, $\ell$. The surface area in terms of these parameters is $2s^2+4\ell s$, while the volume is $\ell s^2$. Therefore, the optimization problem we need to solve is $$ \begin{align} &{\bf \text{maximize: }} & \ell s^2\\ &{\bf \text{subject to:}} & 2s^2+4\ell s=A \end{align} $$ You would then use Lagrange multipliers to solve the optimization problem, finding the largest possible area of such a tent. Repeat the same process for the two given shapes, triangular and semicircular, and then see which of the two gives the largest volume tent.
The Lagrange multiplier equation is $$ \nabla(\ell s^2)=\lambda \nabla (2s^2+4s\ell),\\ (s^2,2s\ell)=\lambda(4s,4s+4\ell) $$ This splits into two equations, $$ s^2=4s\lambda\qquad \text{and}\qquad 2s\ell=\lambda(4s+4\ell) $$ The left equation can be factored in $s$, has two solutions for $s$; either $s=0$, or $s=4\lambda$. The solution $s=0$ is non-physical, so we conclude $\lambda=s/4$. Substituting that into the second equation, we get $$ 2s\ell=(s/4)(4s+4\ell)\quad \implies \quad s^2-s\ell=0 $$ The two solutions are $s=\ell$ and $s=0$. The latter is non-physical, so we conclude $s=\ell$. This is sensible; the largest volume rectangular prism for a given area is the cube!
Finally, we can plug $s=\ell$ into the constraint equation to derive $2s^2+4(s)s=A$, which shows that $s=\sqrt{A/6}$. Therefore, the largest volume tent of this shape has a volume of $s^2\ell=(\sqrt{A/6})^3=(A/6)^{3/2}$.
For your problem, you need to do a very similar process. For example, a semi-circular prism is specified by the radius, $r$, and the height, $h$. You can get two functions for the volume and surface area of such a prism, then do Lagrange multipliers just as I did above.