Answer
Base length $10ft$ and height $5ft$.
Work Step by Step
a. Step 1. Assume the square-shaped base has an edge length of $a$ ft and the height is $h$ ft.
Step 2. The total volume will be $V=a^2h=500ft^3$ and $h=\frac{500}{a^2}$.
Step 3. The total surface area for this open top tank is given by $S=a^2+4ah=a^2+4a\times\frac{500}{a^2}=a^2+\frac{2000}{a}$.
Step 4. Take the derivative to get $S'=2a-\frac{2000}{a^2}$ and letting $S'=0$ to get the extreme, we have $a^3=1000, a=10ft$.
Step 5. This extreme is a minimum because $S'$ changes signs from negative to positive across $a=10$, that is $..(-)..(10)..(+)..$.
Step 6. Thus at $a=10ft, h=5ft$, we have a minimum $S=10^2+4(10)(5)=300ft^2$.
b. The total weight of the metal sheet equals to the product of material density, thickness, and the total surface area. As the density and thickness are constants, we can use the total area to represent the total weight for which we found the minimum amount of the material needed.
