Answer
$x=15ft$, $y=5ft$.
Work Step by Step
a. Step 1.The total volume is given by $V=x^2y=1125ft^3$; we have $y=\frac{1125}{x^2}$.
Step 2. Given the cost equation $c=5(x^2+4xy)+10xy$, we have $c=5(x^2+4x(\frac{1125}{x^2}))+10x(\frac{1125}{x^2})=5(x^2+\frac{4500}{x})+\frac{11250}{x}=5x^2+\frac{33750}{x}$
Step 3. Take the derivative to get $c'=10x-\frac{33750}{x^2}$ and an extreme happens when $c'=0$ which gives $x^3=3375, x=15ft$ and thus $y=\frac{1125}{x^2}=5ft$.
Step 4. The extreme is a minimum because $c'$ changes signs from negative to positive when cross the point $x=15$, that is $..(-)..(15)..(+)..$.
b. The total surface area of the tank is $S=x^2+4xy$ and the cost function becomes $c=5S+10xy$. Clearly, the first term is proportional (factor 5 as unit cost) to the total surface area. The second term is the excavation charge which is proportional (factor 10 as unit cost) to the area of one side of the tank. It is possible to explain the formula in different ways depending on the real life scenarios.
