Chapter 5.3, Problem 36PS

Solution

a.

To determine: the equation that gives the tank’s inner height $h$ in terms of its inner radius $r$ .

The equation is $h=-\frac{r^2\pi +2r\pi +\pi -100}{2r\pi +\pi }$ .

Given information:

The volume of cylindrical concrete water tank is $100$ cubic feet.

Concept Used:

Volume of cylindrical:

The volume of the cylindrical is calculated by $V=\pi r^{2}h$ where $r$ represents the radius of the cylindrical and $h$ represents the height of the cylindrical.

Calculation:

The radius of the inner cylindrical is $r$ .

The radius of the outer cylindrical is $h$ .

The height of the inner cylindrical is $r+1$ .

The height of the outer cylindrical is $h+1$ .

The volume of the inner cylindrical is $V_1=\pi r^{2}h$ .

The volume of the outer cylindrical is $V_0=\pi {\left(r+1\right)}^2\left(h+1\right)$ .

The volume of cylindrical concrete water tank is $V=100$ cubic feet.

The total volume is the difference of the volume of the outside cylindrical and the volume of the inner cylindrical.

  $\begin{array}{c}V=V_0-V_1\\ 100=\pi {\left(r+1\right)}^2\left(h+1\right)-\pi r^{2}h\\ 100=\pi \left(r^2+2r+1\right)\left(h+1\right)-\pi r^{2}h\\ 100=\pi \left(r^{2}h+r^2+2rh+2r+h+1\right)-\pi r^{2}h\\ 100=\pi r^{2}h+\pi r^2+2\pi rh+2\pi r+\pi h+\pi -\pi r^{2}h\\ 100=\pi r^2+2\pi rh+2\pi r+\pi h+\pi \\ 100-\pi r^2-2\pi r-\pi =h\left(2\pi r+\pi \right)\\ h=-\frac{r^2\pi +2r\pi +\pi -100}{2r\pi +\pi }\end{array}$

b.

To determine: the equation that gives the volume $V$ of water that the tank can hold as a function $r$ .

The volume $V$ of water that the tank can hold as a function $r$ is $V=-\frac{r^4\pi +2r^3\pi +r^2\pi -100r^2}{2r+1}$ .

Given information:

The volume of cylindrical concrete water tank is $100$ cubic feet.

Concept Used:

Volume of cylindrical:

The volume of the cylindrical is calculated by $V=\pi r^{2}h$ where $r$ represents the radius of the cylindrical and $h$ represents the height of the cylindrical.

Calculation:

From part(a), the equation is $h=-\frac{r^2\pi +2r\pi +\pi -100}{2r\pi +\pi }$ .

Substitute $-\frac{r^2\pi +2r\pi +\pi -100}{2r\pi +\pi }$ for $h$ in $V=\pi r^{2}h$ .

  $\begin{array}{c}V=\pi r^2\left(-\frac{r^2\pi +2r\pi +\pi -100}{2r\pi +\pi }\right)\\ =r^2\left(-\frac{r^2\pi +2r\pi +\pi -100}{2r+1}\right)\\ =-\frac{r^4\pi +2r^3\pi +r^2\pi -100r^2}{2r+1}\end{array}$

c.

To graph: the equation from part (b) and then determine the value of $r$ and $h$ that maximize the tank’s capacity.

The graph is shown below:

  

The value of $r$ and $h$ that maximize the tank’s capacity is $r=2.742\text{feet}$ and $h=64.946\text{feet}$ .

Given information:

The volume $V$ of water that the tank can hold as a function $r$ is $V=-\frac{r^4\pi +2r^3\pi +r^2\pi -100r^2}{2r+1}$

Concept Used:

Graphing translations of simple rational functions:

To graph a rational function of the form $y=\frac{a}{x-h}+k$ , follow these steps:

• Draw the asymptotes $x=h$ and $y=k$ .
• Plot points to the left and to the right of the vertical asymptote.
• Draw the two branches of the hyperbola so that they pass through the plotted points and approach the asymptotes.

Calculation:

Use the graphing calculator, to graph the function $V=-\frac{r^4\pi +2r^3\pi +r^2\pi -100r^2}{2r+1}$ .

  

From the above graph, it is observed that the radius and height are maximum at $r=2.742$ and $h=64.946$ .