Chapter 4.4, Problem 12E

(a)

To determine

To find: The values of $x$ and $y$ that will minimize the given cost function.

Answer to Problem 12E

  $\text{base}=15\text{ ft}$ and $\text{height}=5\text{ ft}$

Explanation of Solution

Given information: The volume of the open tank is ${\text{1125 ft}}^3$ .

Concept used: The minimum value of the function $f\left(x\right)$ will be at the critical point of that function.

Calculation:

Consider the dimensions of the square base as $x$ by $x$ , and height as $y$ .

From the given information, the volume (V) of the open tank will be as shown below.

  $\begin{array}{c}V=x\left(x\right)\left(y\right)\\ =x^{2}y\end{array}$

Substitute 500 for $V$ into the obtained volume and solve for the expression of $h$ .

  $\begin{array}{c}1125=x^{2}y\\ y=\frac{1125}{x^2}\end{array}$

The total cost of the tank is given by the function, $c=5\left(x^2+4xy\right)+10xy$ .

So,

  $\begin{array}{c}c=5\left(x^2+4xy\right)+10xy\\ =5x^2+30xy\\ =5x^2+30x\left(\frac{1125}{x^2}\right)\\ =5x^2+\left(\frac{33750}{x^2}\right)\end{array}$

Differentiate the obtained function with respect to $x$ .

  $\frac{dc}{dx}=10x-\frac{33750}{x^2}$

Equate the obtained derivative to 0 and solve for $x$ .

  $\begin{array}{c}\frac{dc}{dx}=0\\ 10x-\frac{33750}{x^2}=0\\ x^3=3375\\ x=15\end{array}$

Now, differentiate the equation, $\frac{dc}{dx}=10x-\frac{33750}{x^2}$ , with respect to $x$ .

  $\frac{d^{2}C}{d{x}^2}=10+\frac{67500}{x^3}$

Now, substitute 15 for $x$ into the obtained second derivative and simplify it.

  $\begin{array}{c}{\left.\frac{d^{2}C}{d{x}^2}\right|}_{x=15}=10+\frac{67500}{{15}^3}\\ =30\end{array}$

Since at $x=15$ , $\frac{d^{2}c}{d{x}^2}>0$ , the dimensions of the open tank will be minimum at $x=15$ .

Substitute $x=15$ into the equation, $y=\frac{1125}{x^2}$ , and solve for $y$ .

  $\begin{array}{c}y=\frac{1125}{{15}^2}\\ =5\end{array}$

Therefore, for the given open tank, $\text{base}=15\text{ ft}$ and $\text{height}=5\text{ ft}$.

(b)

To determine

To describe: The possible scenario for the given cost function.

Answer to Problem 12E

Cost of the material for the open tank is $5{\text{ dollar/ft}}^2$

Excavation charge is $10{\text{ dollar/ft}}^2$

Explanation of Solution

Given information: The volume of the open tank is ${\text{1125 ft}}^3$ and the cost function is $c=5\left(x^2+4xy\right)+10xy$ . Also, the excavation charge is the product of $x$ and $y$ .

The cost of the material for the open tank is calculated on the basis of the surface area of the tank, that is, $x^2+4xy$ . So, this implies that the cost of the material for the open tank is $5{\text{ dollar/ft}}^2$ .

It is given that the excavation charge is the product of $x$ and $y$ . So, according to the given statement, the excavation charge is $10{\text{ dollar/ft}}^2$ .