Minimizing the surface area of a rectangular prism with volume $50000$L. A town wants to build a water tank to hold 50000L of water. They want to minimize the amount of material used to build the tank by minimizing the surface area. If they make the water tank into a closed rectangular prism, what would the dimensions be?
Volume of the cube is $50 kL$. Hence, $lbh=50000$. If the surface area is minimum, then $lb+2h(l+b)$ has to be minimized.
Using  $h=\frac{50000}{lb}$ we get $SA=lb+100000\frac{l+b}{lb}$.
By argument of symmetry, $l=b$.
Minimizing $l^2+\frac{200000}{l}$ gives $l=10\sqrt[3]{100}$ and $h=5\sqrt[3]{100}$
A cube, $a=\sqrt[3]{50}\approx3.68 m=368 cm$, $A=6a^²≈81.22 m^²$
If a shape of water tank is arbitrary, then it's a sphere , $R=\sqrt[3]{\frac{3*50}{4\pi})}\approx2.28 m, A_1=4\pi R^2\approx 65.52 m^2$