3.3 Inequalities involving Rational Functions and Applications

In this section, we solve equations and inequalities involving rational functions and explore associated application problems. Our first example showcases the critical difference in procedure between solving equations and inequalities.

Example 1

Solve $\frac{x^{3} - 2 x + 1}{x - 1} = \frac{1}{2} x - 1$ .
Solve $\frac{x^{3} - 2 x + 1}{x - 1} \geq \frac{1}{2} x - 1$ .
Verify your solutions using a graphing utility.

To solve the equation, we clear denominators
$\begin{matrix} \frac{x^{3} - 2 x + 1}{x - 1} & = & \frac{1}{2} x - 1 \\ (\frac{x^{3} - 2 x + 1}{x - 1}) \cdot 2 (x - 1) & = & (\frac{1}{2} x - 1) \cdot 2 (x - 1) \\ 2 x^{3} - 4 x + 2 & = & x^{2} - 3 x + 2 & expand \\ 2 x^{3} - x^{2} - x & = & 0 \\ x (2 x + 1) (x - 1) & = & 0 & factor \\ x & = & - \frac{1}{2}, 0, 1 \end{matrix}$
Since we cleared denominators, we need to check for extraneous solutions. Sure enough, we see that $x equals 1$ does not satisfy the original equation, so our only solutions are $x = - \frac{1}{2}$ and $x equals 0$ .
To solve the inequality, it may be tempting to begin as we did with the equation namely by multiplying both sides by the quantity $open parenthesis x minus 1 close parenthesis$ . The problem is that, depending on $x$ , $open parenthesis x minus 1 close parenthesis$ may be positive (which doesn't affect the inequality) or $open parenthesis x minus 1 close parenthesis$ could be negative (which would reverse the inequality). Instead of working by cases, we collect all of the terms on one side of the inequality with $0$ on the other and make a sign diagram using the technique given on page 3.2 in Section 3.2.
$\begin{matrix} \frac{x^{3} - 2 x + 1}{x - 1} & \geq & \frac{1}{2} x - 1 \\ \frac{x^{3} - 2 x + 1}{x - 1} - \frac{1}{2} x + 1 & \geq & 0 \\ \frac{2 (x^{3} - 2 x + 1)}{2 (x - 1)} - \frac{x (x - 1)}{2 (x - 1)} + \frac{2 (x - 1)}{2 (x - 1)} & \geq & 0 & get a common denominator \\ \frac{2 (x^{3} - 2 x + 1) - x (x - 1) + 2 (x - 1)}{2 (x - 1)} & \geq & 0 \\ \frac{2 x^{3} - x^{2} - x}{2 x - 2} & \geq & 0 & expand \end{matrix}$
Viewing the left hand side as a rational function $r (x)$ we make a sign diagram. The only value excluded from the domain of $r$ is $x equals 1$ which is the solution to $2 x - 2 = 0$ . The zeros of $r$ are the solutions to $2 x^{3} - x^{2} - x = 0$ , which we have already found to be $x equals 0$ , $x = - \frac{1}{2}$ and $x equals 1$ , the latter was discounted as a zero because it is not in the domain. Choosing test values in each test interval, we construct the sign diagram below.
Figure 3.82
We are interested in where $r ⁢ open parenthesis x close parenthesis greater than or equal to 0$ . We see $r ⁢ open parenthesis x close parenthesis greater than 0$ , or $(+)$ , on the intervals $(- \infty, - \frac{1}{2})$ , $open parenthesis 0, 1 close parenthesis$ and $open parenthesis 1, ∞ close parenthesis$ . We know $r ⁢ open parenthesis x close parenthesis equals 0$ when $x = - \frac{1}{2}$ and $x equals 0$ . Hence, $r ⁢ open parenthesis x close parenthesis greater than or equal to 0$ on $(- \infty, - \frac{1}{2}] \cup [0, 1) \cup (1, \infty)$ .
To check our answers graphically, let $f (x) = \frac{x^{3} - 2 x + 1}{x - 1}$ and $g (x) = \frac{1}{2} x - 1$ . The solutions to $f (x) = g (x)$ are the $x$ -coordinates of the points where the graphs of $y = f (x)$ and $y = g (x)$ intersect. We graph both $f$ and $g$ below (the graph of $g$ is the line and is slightly lighter in color.) We find only two intersection points, $(- 0.5, - 1.25)$ and $open parenthesis 0, minus 1 close parenthesis$ which correspond to our solutions $x = - \frac{1}{2}$ and $x equals 0$ . The solution to $f (x) \geq g (x)$ represents not only where the graphs meet, but the intervals over which the graph of $y = f (x)$ is above ( $>$ ) the graph of $g (x)$ . From the graph, this appears to happen on on $(- \infty, - \frac{1}{2}] \cup [0, \infty)$ which almost matches the answer we found analytically. We have to remember that $f$ is not defined at $x equals 1$ , so it cannot be included in our solution.¹
Figure 3.83
Figure 3.84

The important take-away from Example 3.3.1 is not to clear fractions when working with an inequality unless you know for certain the sign of the denominators. We offer another example.

Example 2

Solve: $2 t {(3 t - 2)}^{- 1} \leq 3 t^{2} {(3 t - 2)}^{- 2}$ . Check your answer using a graphing utility.

We begin by rewriting the terns with negative exponents as fractions and gathering all nonzero terms to one side of the inequality:

$\begin{matrix} 2 t {(3 t - 2)}^{- 1} & \leq & 3 t^{2} {(3 t - 2)}^{- 2} \\ \frac{2 t}{3 t - 2} & \leq & \frac{3 t^{2}}{{(3 t - 2)}^{2}} \\ \frac{2 t}{3 t - 2} - \frac{3 t^{2}}{{(3 t - 2)}^{2}} & \leq & 0 \\ \frac{2 t (3 t - 2)}{{(3 t - 2)}^{2}} - \frac{3 t^{2}}{{(3 t - 2)}^{2}} & \leq & 0 & get a common denominator \\ \frac{2 t (3 t - 2) - 3 t^{2}}{{(3 t - 2)}^{2}} & \leq & 0 \\ \frac{3 t^{2} - 4 t}{{(3 t - 2)}^{2}} & \leq & 0 & expand \end{matrix}$

We define $r (t) = \frac{3 t^{2} - 4 t}{{(3 t - 2)}^{2}}$ and set about constructing a sign diagram for $r$ . Solving ${(3 t - 2)}^{2} = 0$ gives $t = \frac{2}{3}$ , our sole excluded value. To find the zeros of $r$ , we set $r (t) = \frac{3 t^{2} - 4 t}{{(3 t - 2)}^{2}} = 0$ and solve $3 t^{2} - 4 t = 0$ . Factoring gives $t (3 t - 4) = 0$ so our solutions are $t equals 0$ and $t = \frac{4}{3}$ . After choosing test values, we get the sign diagram below on the left. Since we are looking for where $r ⁢ open parenthesis t close parenthesis less than or equal to 0$ , we are looking for where $r (t)$ is $(-)$ or $r ⁢ open parenthesis t close parenthesis equals 0$ . Hence, our final answer is $[0, \frac{2}{3}) \cup (\frac{2}{3}, \frac{4}{3}]$ . Below on the right, we graph $f (t) = 2 t {(3 t - 1)}^{- 1}$ (the darker curve), $g (t) = 3 t^{2} {(3 t - 2)}^{- 2}$ , and vertical asymptote $x = \frac{2}{3}$ , the dashed line. Sure enough, the graph of $f$ intersects the graph of $g$ when $t equals 0$ and $t = \frac{4}{3}$ . Moreover, the graph of $f$ is below the graph of $g$ everywhere they are defined between these values, in accordance with our algebraic solution.

Figure 3.85

Figure 3.86

One thing to note about Example 3.3.2 is that the quantity $open parenthesis 3 ⁢ t minus 2 close parenthesis squared greater than or equal to 0$ for all values of $t$ . Hence, as long as we remember $t = \frac{2}{3}$ is excluded from consideration, we could actually multiply both sides of the inequality in Example 3.3.2 by $open parenthesis 3 ⁢ t minus 2 close parenthesis squared$ to obtain $2 ⁢ t ⁢ open parenthesis 3 ⁢ t minus 2 close parenthesis less than or equal to 3 ⁢ t squared$ . We could then solve this (slightly easier) inequality using the methods of Section 1.4 as long as we remember to exclude $t = \frac{2}{3}$ from our solution. Once again, the more you understand, the less you have to memorize. If you know the `why' behind an algorithm instead of just the `how,' you will know when you can short-cut it.

Our next example is an application of average cost. Recall from Definition 3.8 if $C (x)$ represents the cost to make $x$ items then the average cost per item is given by $\bar{C} (x) = \frac{C (x)}{x}$ , for $x greater than 0$ .

Example 3

Recall from Example 1.2.3 that the cost, $C (x)$ , in dollars, to produce $x$ PortaBoy game systems for a local retailer is $C (x) = 80 x + 150$ , $x greater than or equal to 0$ .

Find an expression for the average cost function, $\bar{C} (x)$ .
Solve $C bar ⁢ open parenthesis x close parenthesis less than 100$ and interpret.
Find and interpret $lim x → ∞ C bar ⁢ open parenthesis x close parenthesis$ .

From $\bar{C} (x) = \frac{C (x)}{x}$ , we obtain $\bar{C} (x) = \frac{80 x + 150}{x}$ . The domain of $C$ is $x greater than or equal to 0$ , but since $x equals 0$ causes problems for $\bar{C} (x)$ , we get our domain to be $x greater than 0$ , or $open parenthesis 0, ∞ close parenthesis$ .
Solving $C bar ⁢ open parenthesis x close parenthesis less than 100$ means we solve $\frac{80 x + 150}{x} < 100$ . We proceed as in the previous example.
$\begin{matrix} \frac{80 x + 150}{x} & < & 100 \\ \frac{80 x + 150}{x} - 100 & < & 0 \\ \frac{80 x + 150 - 100 x}{x} & < & 0 & common denominator \\ \frac{150 - 20 x}{x} & < & 0 \end{matrix}$
If we take the left hand side to be a rational function $r (x)$ , we need to keep in mind that the applied domain of the problem is $x greater than 0$ . This means we consider only the positive half of the number line for our sign diagram. On $open parenthesis 0, ∞ close parenthesis$ , $r$ is defined everywhere so we need only look for zeros of $r$ . Setting $r ⁢ open parenthesis x close parenthesis equals 0$ gives $150 - 20 x = 0$ , so that $x = \frac{15}{2} = 7.5$ . The test intervals on our domain are $open parenthesis 0, 7.5 close parenthesis$ and $open parenthesis 7.5, ∞ close parenthesis$ . We find $r ⁢ open parenthesis x close parenthesis less than 0$ on $open parenthesis 7.5, ∞ close parenthesis$ .
Figure 3.87
In the context of the problem, $x$ represents the number of PortaBoy games systems produced and $\bar{C} (x)$ is the average cost to produce each system. Solving $C bar ⁢ open parenthesis x close parenthesis less than 100$ means we are trying to find how many systems we need to produce so that the average cost is less than $$ 100$ per system. Our solution, $open parenthesis 7.5, ∞ close parenthesis$ tells us that we need to produce more than $7.5$ systems to achieve this. Since it doesn't make sense to produce half a system, our final answer is $open bracket 8, ∞ close parenthesis$ .
To find $lim x → ∞ C bar ⁢ open parenthesis x close parenthesis$ , we note that we can rewrite $\bar{C} (x) = \frac{80 x + 150}{x} = 80 + \frac{150}{x}$ . As $x → ∞$ , note that $\frac{150}{x} \to 0$ so $lim_{x \to \infty} \bar{C} (x) = 80 + 0 = 80$ . Thus the average cost per system is getting closer to $$ 80$ per system. Since $\frac{150}{x} > 0$ for all $x greater than 0$ , we have that $C bar ⁢ open parenthesis x close parenthesis greater than 80$ for all $x greater than 0$ . This means that the average cost per system is always greater than $$ 80$ per system, but the average cost is approaching this amount as more and more systems are produced. Looking back at Example 1.2.3, we realize $$ 80$ is the variable cost per system the cost per system above and beyond the fixed initial cost of $$ 150$ . Another way to interpret our answer is that `infinitely' many systems would need to be produced to effectively `zero out' the fixed cost.

Note that number in Example 3.3.3 is another opportunity to short-cut the standard algorithm and obtain the solution more quickly if we take stock of the situation. Since the applied domain is $x greater than 0$ , we can multiply through the inequality $\frac{80 x + 150}{x} < 100$ by $x$ without worrying about changing the sense of the inequality. This reduces the problem to $80 ⁢ x plus 150 less than 100 ⁢ x$ , a basic linear inequality whose solution is readily seen to be $x greater than 7.5$ . It is absolutely critical here that $x greater than 0$ . Indeed, any time you decide to multiply an inequality by a variable expression, it is necessary to justify why the inequality is preserved. Our next example is another classic `box with no top' problem. The reader is encouraged to compare and contrast this problem with Example 2.1.4 in Section 2.1.

Example 4

A box with a square base and no top is to be constructed so that it has a volume of $1000$ cubic centimeters. Let $x$ denote the width of the box, in centimeters as seen below.

Figure 3.88

Explain why the height of the box (in centimeters) is a function of the width $x$ . Call this function $h$ and find an expression for $h (x)$ , complete with an appropriate applied domain.
Solve $h (x) \geq x$ and interpret.
Find and interpret $lim_{x \to 0^{+}} h (x)$ and $lim x → ∞ h ⁢ open parenthesis x close parenthesis$ .
Express the surface area of the box as a function of $x$ , $S (x)$ and state the applied domain.
Use a graphing utility to approximate (to two decimal places) the dimensions of the box which minimize the surface area.

We are told that the volume of the box is $1000$ cubic centimeters and that $x$ represents the width, in centimeters. Since $x$ represents a physical dimension of a box, we have that $x greater than 0$ . From geometry, we know $volume = width \times height \times depth$ . Since the base of the box is a square, the width and the depth are both $x$ centimeters. Hence, $1000 equals x squared ⁢ open parenthesis height close parenthesis$ . Solving for the height, we get $height = \frac{1000}{x^{2}}$ . In other words, for each width $x greater than 0$ , we are able to compute the ² corresponding height using the formula $\frac{1000}{x^{2}}$ . Hence, the height is a function of $x$ . Using function notation, we write $h (x) = \frac{1000}{x^{2}}$ . As mentioned before, our only restriction is $x greater than 0$ so the domain of $h$ is $open parenthesis 0, ∞ close parenthesis$ .
To solve $h (x) \geq x$ , we proceed as before and collect all nonzero terms on one side of the inequality in order to use a sign diagram.
$\begin{matrix} h (x) & \geq & x \\ \frac{1000}{x^{2}} & \geq & x \\ \frac{1000}{x^{2}} - x & \geq & 0 \\ \frac{1000 - x^{3}}{x^{2}} & \geq & 0 & common denominator \end{matrix}$
We consider the left hand side of the inequality as our rational function $r (x)$ . We see immediately the only value excluded from the domain of $r$ is $0$ , but since our applied domain is $x greater than 0$ , we restrict our attention to the interval $open parenthesis 0, ∞ close parenthesis$ . The sole zero of $r$ comes when $1000 - x^{3} = 0$ , or when $x equals 10$ . Choosing test values in the intervals $open parenthesis 0, 10 close parenthesis$ and $open parenthesis 10, ∞ close parenthesis$ gives the following:
Figure 3.89
We see $r ⁢ open parenthesis x close parenthesis greater than 0$ on $open parenthesis 0, 10 close parenthesis$ , and since $r ⁢ open parenthesis x close parenthesis equals 0$ at $x equals 10$ , our solution is $open parenthesis 0, 10 close bracket$ . In the context of the problem, $h (x)$ represents the height of the box while $x$ represents the width (and depth) of the box. Solving $h (x) \geq x$ is tantamount to finding the values of $x$ which result in a box where the height is at least as big as the width (and, in this case, depth.) Our answer tells us the width of the box can be at most $10$ centimeters for this to happen.³
Since $h (x) = \frac{1000}{x^{2}}$ , $lim_{x \to 0^{+}} h (x) = \infty$ . This means that the smaller the width $x$ (and, in this case, depth), the larger the height $h$ has to be in order to maintain a volume of $1000$ cubic centimeters. On the other hand, $lim x → ∞ h ⁢ open parenthesis x close parenthesis equals 0$ , which means that in order to maintain a volume of $1000$ cubic centimeters, the width and depth must get bigger as the height becomes smaller.
Since the box has no top, the surface area can be found by adding the area of each of the sides to the area of the base. The base is a square of dimensions $x$ by $x$ , and each side has dimensions $x$ by $h (x)$ . We get the surface area, $S (x) = x^{2} + 4 x h (x)$ . Since $h (x) = \frac{1000}{x^{2}}$ , we have $S (x) = x^{2} + 4 x (\frac{1000}{x^{2}}) = x^{2} + \frac{4000}{x}$ . The domain of $S$ is the same as $h$ , namely $open parenthesis 0, ∞ close parenthesis$ , for the same reasons as above.
To graph $y = S (x)$ , we create a table of values to help us define a good viewing window. Doing so, we find a local minim when $x approximately equals 12.60$ . As far as we can tell,⁴ this is the only local extremum, so it is the (absolute) minimum as well. This means that the width and depth of the box should each measure approximately $12.60$ centimeters. To determine the height, we find $h ⁢ open parenthesis 12.60 close parenthesis approximately equals 6.30$ , so the height of the box should be approximately $6.30$ centimeters.⁵
Figure 3.90

Our last example uses regression to verify a very famous scientific law.

Example 5

Boyle's Law states that when temperature is held constant, the pressure of a gas is inversely proportional to the volume of the gas.⁶ According to this website the actual data relating the volume $V$ of a gas and its pressure $P$ used by Boyle and his assistant in 1662 to formulate this law is given below. (NOTE: both pressure and volume here are given in `arbitrary units.')

$V, 48, 46, 44, 42, 40, 38, 36, 34, 32, 30, 28, 26, 24; P, 29.13, 30.56, 31.94, 33.5, 35.31, 37, 39.31, 41.63, 44.19, 47.06, 50.31, 54.31, 58.81$

$V, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12; P, 61.31, 64.06, 67.06, 70.69, 74.13, 77.88, 82.75, 87.88, 93.06, 100.44, 107.81, 117.56$

Assuming $P$ and $V$ are inversely proportional, estimate the constant of proportionality, $k$ .
Use a graphing utility to fit a curve of the form $P = \frac{k}{V}$ to these data.

Recall if $P$ and $V$ are inversely proportional, there is a real number $k$ so $P V = k$ for all values of $P$ and $V$ . Multiplying the corresponding $P$ and $V$ values from the data together result in numbers which are consistently approximately $1400$ . This gives us confidence in the claim $P$ and $V$ are inversely proportional and suggests $k approximately equals 1400$ .
We plot the pairs $(V, P)$ and run a regression, the results of which are below. To our amazement, the graphing utility reports $k approximately equals 1406.9$ with $R squared approximately equals 1$ . This means the data are a very good fit to the model $P = \frac{k}{V}$ , or $P V = k$ , hence verifying Boyle's Law for this set of data.
Figure 3.91

Exercises

(Review of Solving Equations):⁷ In Exercises -, solve the rational equation. Be sure to check for extraneous solutions.

$\frac{x}{5 x + 4} = 3$
Answer
$x = - \frac{6}{7}$
$\frac{3 x - 1}{x^{2} + 1} = 1$
Answer
$x = 1, x = 2$
$\frac{1}{t + 3} + \frac{1}{t - 3} = \frac{t^{2} - 3}{t^{2} - 9}$
Answer
$t = - 1$
$\frac{2 t + 17}{t + 1} = t + 5$
Answer
$t = - 6, x = 2$
$\frac{z^{2} - 2 z + 1}{z^{3} + z^{2} - 2 z} = 1$
Answer
No solution
$\frac{4 z - z^{3}}{z^{2} - 9} = 4 z$
Answer
$z = 0, z = \pm 2 \sqrt{2}$

In Exercises -, solve the rational inequality. Express your answer using interval notation.

$\frac{1}{x + 2} \geq 0$
Answer
$open parenthesis minus 2, ∞ close parenthesis$
$\frac{5}{x + 2} \geq 1$
Answer
$open parenthesis minus 2, 3 close bracket$
$\frac{x}{x^{2} - 1} < 0$
Answer
$(- \infty, - 1) \cup (0, 1)$
$\frac{4 t}{t^{2} + 4} \geq 0$
Answer
$open bracket 0, ∞ close parenthesis$
$\frac{2 t + 6}{t^{2} + t - 6} < 1$
Answer
$(- \infty, - 3) \cup (- 3, 2) \cup (4, \infty)$
$\frac{5}{t - 3} + 9 < \frac{20}{t + 3}$
Answer
$(- 3, - \frac{1}{3}) \cup (2, 3)$
$\frac{6 z + 6}{2 + z - z^{2}} \leq z + 3$
Answer
$open parenthesis minus 1, 0 close bracket union open parenthesis 2, ∞ close parenthesis$
$\frac{6}{z - 1} + 1 > \frac{1}{z + 1}$
Answer
$(- \infty, - 3) \cup (- 2, - 1) \cup (1, \infty)$
$\frac{3 z - 1}{z^{2} + 1} \leq 1$
Answer
$open parenthesis minus ∞, 1 close bracket union open bracket 2, ∞ close parenthesis$
$(2 x + 17) {(x + 1)}^{- 1} > x + 5$
Answer
$(- \infty, - 6) \cup (- 1, 2)$
$(4 x - x^{3}) {(x^{2} - 9)}^{- 1} \geq 4 x$
Answer
$(- \infty, - 3) \cup [- 2 \sqrt{2}, 0] \cup [2 \sqrt{2}, 3)$
${(x^{2} + 1)}^{- 1} < 0$
Answer
No solution
$(2 t - 8) {(t + 1)}^{- 1} \leq (t^{2} - 8 t) {(t + 1)}^{- 2}$
Answer
$[- 4, - 1) \cup (- 1, 2]$
$(t - 3) (2 t + 7) {(t^{2} + 7 t + 6)}^{- 2} \geq {(t^{2} + 7 t + 6)}^{- 1}$
Answer
$(- \infty, - 6) \cup (- 6, - 3] \cup [9, \infty)$
$60 z^{- 2} + 23 z^{- 1} \geq 7 {(z - 4)}^{- 1}$
Answer
$open bracket minus 3, 0 close parenthesis union open parenthesis 0, 4 close parenthesis union open bracket 5, ∞ close parenthesis$
$2 z + 6 {(z - 1)}^{- 1} \geq 11 - 8 {(z + 1)}^{- 1}$
Answer
$(- 1, - \frac{1}{2}] \cup (1, \infty)$

In Exercises -, use the the graph of the given rational function to solve the stated inequality.

Solve $f ⁢ open parenthesis x close parenthesis greater than or equal to 0$ .
Figure 3.92 $y = f (x)$ , asymptotes: $x equals 0$ , $y equals 1$ .
Answer
$f ⁢ open parenthesis x close parenthesis greater than or equal to 0$ on $open parenthesis minus ∞, 0 close parenthesis union open bracket 3, ∞ close parenthesis$ .
Solve $f ⁢ open parenthesis x close parenthesis less than 1$ .
Figure 3.93 $y = f (x)$ , asymptotes: $x equals 0$ , $y equals 1$ .
Answer
$f ⁢ open parenthesis x close parenthesis less than 1$ on $open parenthesis 0, ∞ close parenthesis$ .
Solve $g ⁢ open parenthesis t close parenthesis greater than or equal to minus 1$ .
Figure 3.94 $y = g (t)$ , asymptotes: $t equals 2$ , $y equals 0$ .
Answer
$g ⁢ open parenthesis t close parenthesis greater than or equal to minus 1$ on $open parenthesis minus ∞, 1 close bracket union open parenthesis 2, ∞ close parenthesis$ .
Solve $minus 1 less than or equal to g ⁢ open parenthesis t close parenthesis less than 1$ .
Figure 3.95 $y = g (t)$ , asymptotes: $t equals 2$ , $y equals 0$ .
Answer
$minus 1 less than or equal to g ⁢ open parenthesis t close parenthesis less than 1$ on $open parenthesis minus ∞, 1 close bracket union open parenthesis 3, ∞ close parenthesis$ .
Solve $r ⁢ open parenthesis z close parenthesis less than or equal to 1$
Figure 3.96 $y = r (z)$ , asymptotes: $z equals 0$ , $y equals 0$ .
Answer
$r ⁢ open parenthesis z close parenthesis less than or equal to 1$ on $(- \infty, - 1] \cup (1, \infty)$ .
Solve $r ⁢ open parenthesis z close parenthesis greater than 0$ .
Figure 3.97 $y = r (z)$ , asymptotes: $z equals 0$ , $y equals 0$ .
Answer
$r ⁢ open parenthesis z close parenthesis greater than 0$ on $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, 1 close parenthesis union open parenthesis 1, ∞ close parenthesis$ .
In Exercise in Section 2.1, the function $C (x) = .03 x^{3} - 4.5 x^{2} + 225 x + 250$ , for $x greater than or equal to 0$ was used to model the cost (in dollars) to produce $x$ PortaBoy game systems. Using this cost function, find the number of PortaBoys which should be produced to minimize the average cost $\bar{C}$ . Round your answer to the nearest number of systems.
Answer
The absolute minimum of $y = \bar{C} (x)$ occurs at $approximately equals open parenthesis 75.73, 59.57 close parenthesis$ . Since $x$ represents the number of game systems, we check $C bar ⁢ open parenthesis 75 close parenthesis approximately equals 59.58$ and $C bar ⁢ open parenthesis 76 close parenthesis approximately equals 59.57$ . Hence, to minimize the average cost, $76$ systems should be produced at an average cost of $$ 59.57$ per system.
Suppose we are in the same situation as Example 3.3.4. If the volume of the box is to be $500$ cubic centimeters, use a graphing utility to find the dimensions of the box which minimize the surface area. What is the minimum surface area? Round your answers to two decimal places.
Answer
The width (and depth) should be $10.00$ centimeters, the height should be $5.00$ centimeters. The minimum surface area is $300.00$ square centimeters.
The box for the new Sasquatch-themed cereal, `Crypt-Os', is to have a volume of $140$ cubic inches. For aesthetic reasons, the height of the box needs to be $1.62$ times the width of the base of the box.⁸ Find the dimensions of the box which will minimize the surface area of the box. What is the minimum surface area? Round your answers to two decimal places.
Answer
The width of the base of the box should be approximately $4.12$ inches, the height of the box should be approximately $6.67$ inches, and the depth of the base of the box should be approximately $5.09$ inches. The minimum surface area is approximately $164.91$ square inches.
Sally is Skippy's neighbor from Exercise in Section 1.4. Sally also wants to plant a vegetable garden along the side of her home. She doesn't have any fencing, but wants to keep the size of the garden to 100 square feet. What are the dimensions of the garden which will minimize the amount of fencing she needs to buy? What is the minimum amount of fencing she needs to buy? Round your answers to the nearest foot. (Note: Since one side of the garden will border the house, Sally doesn't need fencing along that side.)
Answer
The dimensions are approximately $7$ feet by $14$ feet. Hence, the minimum amount of fencing required is approximately $28$ feet.
Another Classic Problem: A can is made in the shape of a right circular cylinder and is to hold one pint. (For dry goods, one pint is equal to $33.6$ cubic inches.)⁹
1. Find an expression for the volume $V$ of the can in terms of the height $h$ and the base radius $r$ .
2. Find an expression for the surface area $S$ of the can in terms of the height $h$ and the base radius $r$ . (Hint: The top and bottom of the can are circles of radius $r$ and the side of the can is really just a rectangle that has been bent into a cylinder.)
3. Using the fact that $V equals 33.6$ , write $S$ as a function of $r$ and state its applied domain.
4. Use your graphing calculator to find the dimensions of the can which has minimal surface area.
Answer
1. $V = π r^{2} h$
2. $S = 2 π r^{2} + 2 π r h$
1. $S (r) = 2 π r^{2} + \frac{67.2}{r},$ Domain $r greater than 0$
2. $r approximately equals 1.749$ in. and $h approximately equals 3.498$ in.
A right cylindrical drum is to hold 7.35 cubic feet of liquid. Find the dimensions (radius of the base and height) of the drum which would minimize the surface area. What is the minimum surface area? Round your answers to two decimal places.
Answer
The radius of the drum should be approximately $1.05$ feet and the height of the drum should be approximately $2.12$ feet. The minimum surface area of the drum is approximately $20.93$ cubic feet.
In Exercise in Section 3.1, the population of Sasquatch in Portage County is modeled by
$P (t) = \frac{150 t}{t + 15}, t \geq 0,$
where $t equals 0$ corresponds to the year 1803. According to this model, when were there fewer than 100 Sasquatch in Portage County?
Answer
$P ⁢ open parenthesis t close parenthesis less than 100$ on $open parenthesis minus 15, 30 close parenthesis$ , and the portion of this which lies in the applied domain is $open bracket 0, 30 close parenthesis$ . Since $t equals 0$ corresponds to the year 1803, from 1803 through the end of 1832, there were fewer than 100 Sasquatch in Portage County.

We invite the reader to show there is a hole in the graph of $y = f (x)$ at $open parenthesis 1, 1 close parenthesis$ . ↩
that is, the one and only one ↩
As with the previous example, knowing $x greater than 0$ means $x squared greater than 0$ so we can clear denominators right away and solve $x cubed less than or equal to 1000$ , or $x less than or equal to 10$ . Coupled with our applied domain, $x greater than 0$ , we would arrive at the same solution, $open parenthesis 0, 10 close bracket$ . ↩
without Calculus, that is... ↩
The $y$ -coordinate here, $476.22$ means the minimum surface area possible is $476.22$ square centimeters. Minimizing the surface area minimizes the material required to make the box, therein helping to reduce the cost of the box. ↩
For a review of what this means, see Section A.14. ↩
For more review, see Section A.12. ↩
1.62 is a crude approximation of the so-called `Golden Ratio' $ϕ = \frac{1 + \sqrt{5}}{2}$ . ↩
According to www.dictionary.com , there are different values given for this conversion. We use $33.6 ⁢ in cubed$ for this problem. ↩

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.