Tent - spherical cap

I have a tent in the shape of a spherical cap. Assume we want the volume to be 4 cubic meters, to sleep two or three people. Assume that the material making up the dome of the ten is twice as expensive per square as the material touching the ground. What should the dimensions of the tent be so that cost of materials is minimum? S=2*PI*R*h and
V=(1/3)*PI*h*h*(3R-h).

Thank you!

Final Answer:

S =  9.672 m²
h =  1.2407 m
R =  1.2407 m

Step-by-step explanation:

$$\begin{gathered} S=2\pi \cdot R\cdot h \\ V=4{\text{m}}^3 \\ V=\frac{1}{3}\cdot \pi \cdot h\cdot h\cdot (3R-h) \\ 3\cdot V/\pi =h^2\cdot (3R-h) \\ 3\cdot V/\pi /h^2=3R-h \\ 3\cdot V/\pi /h^2+h=3R \\ R=\frac{V}{\pi \cdot h^2}+h/3 \\ S=2\pi \cdot R\cdot h \\ S=2\pi \cdot (\frac{V}{\pi \cdot h^2}+h/3)\cdot h \\ S=(2\pi h^2)/3+(2V)/h \\ {S}^{\prime }=dS/dh=(4\pi h)/3-(2V)/h^2 \\ h=\sqrt[3]{6/\pi }=\sqrt[3]{6/3.1416}\doteq 1.2407\text{m} \\ R=\frac{V}{\pi \cdot h^2}+h/3=\frac{4}{3.1416\cdot 1.2407^2}+1.2407/3\doteq 1.2407\text{m} \\ S=2\pi \cdot R\cdot h=2\cdot 3.1416\cdot 1.2407\cdot 1.2407=9.672{\text{m}}^2 \end{gathered}$$

$$\begin{gathered} R=1.2407\doteq 1.2407\text{m} \\ \text{ Verifying Solution: } \\ {V}_2=\frac{1}{3}\cdot \pi \cdot h^2\cdot (3\cdot R-h)=\frac{1}{3}\cdot 3.1416\cdot 1.2407^2\cdot (3\cdot 1.2407-1.2407)=4{\text{m}}^3 \end{gathered}$$

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