Chapter 4.4, Problem 11E

(a)

To determine

To find: The dimensions for the base and height that will make the tank weigh as little as possible.

Answer to Problem 11E

  $\text{base}=10\text{ ft}$ and $\text{height}=5\text{ ft}$

Explanation of Solution

Given information: The volume of the open tank is $500{\text{ ft}}^3$ .

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 $h$ .

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

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

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

  $\begin{array}{c}500=x^{2}h\\ h=\frac{500}{x^2}\end{array}$

Now, write the function of a surface area (S) for the open tank.

  $\begin{array}{c}S=x^2+\left(4xh\right)\\ =x^2+4x\left(\frac{500}{x^2}\right)\\ =x^2+2000x^{-1}\end{array}$

Differentiate the obtained function with respect to $x$ .

  $\frac{dS}{dx}=2x-\frac{2000}{x^2}$

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

  $\begin{array}{c}\frac{dS}{dx}=0\\ 2x-\frac{2000}{x^2}=0\\ 2x=\frac{2000}{x^2}\\ x^3=1000\\ x=10\end{array}$

Now, differentiate the equation, $\frac{dS}{dx}=2x-\frac{2000}{x^2}$ , with respect to $x$ .

  $\frac{d^{2}S}{d{x}^2}=2+\frac{4000}{x^3}$

Now, substitute 10 for $x$ into the obtained second derivative and simplify it.

  $\begin{array}{c}{\left.\frac{d^{2}S}{d{x}^2}\right|}_{x=10}=2+\frac{4000}{{10}^3}\\ =6\end{array}$

Since at $x=10$ , $\frac{d^{2}S}{d{x}^2}>0$ , the dimensions of the open tank will be minimum at $x=10$ .

Substitute $x=10$ into the equation, $h=\frac{500}{x^2}$ , and solve for $h$ .

  $\begin{array}{c}h=\frac{500}{{10}^2}\\ =5\end{array}$

Therefore, for the given open tank, $\text{base}=10\text{ ft}$ and $\text{height}=5\text{ ft}$.

(b)

To determine

To describe: How would you took weight into account?

Answer to Problem 11E

To have a minimum weigh of the given tank, the dimensions of the tank should be minimum.

Explanation of Solution

Given information: The volume of the open tank is $500{\text{ ft}}^3$ .

In order to have a minimum weigh of the given tank, the dimensions of the tank should be minimum. For the minimum weight of the given open tank, the four sides should have minimum surface area.