Optimization with Rational and Exponential Models

On the optimization page, most of the objective functions were polynomials: areas of fields, volumes of boxes. Many real models aren’t polynomials. The cost per item of making $x$ items is a rational function. Ticket sales that drop by a fixed percentage for each dollar of price follow an exponential model, and so does the amount of a medicine left in the body. The method is the same, but the algebra changes, and the domains are often open-ended, so justifying your answer takes a little more care.

Key ideas

The same steps, new models

The setup from the optimization page still applies: define variables, write the objective as a function of one variable, state the domain, find critical points, justify, and answer the question in context. What’s new is the kind of function you’ll differentiate:

• Rational models like $\dfrac{2000}{x} + 15 + 0.05x$ or $\dfrac{40t}{(t + 9)(t + 3)}$, using the quotient rule or a negative exponent.
• Exponential models like $5000pe^{-0.02p}$ or $12te^{-0.4t}$, using the product rule and the rules from derivatives of exponential functions.

Solving f′(x) = 0 for exponential models

A product like $5000pe^{-0.02p}$ has derivative

$$5000e^{-0.02p} + 5000p\big(-0.02e^{-0.02p}\big) = 5000e^{-0.02p}(1 - 0.02p)$$

Factor out the exponential. Since $e^{\text{anything}} \gt 0$, it can never be $0$, so the derivative is $0$ only when the other factor is: $1 - 0.02p = 0$. The same works for $b^{-p}$ with any base $b \gt 0$.

Solving f′(x) = 0 for rational models

For a quotient $\dfrac{u}{v}$, the derivative is $\dfrac{u'v - uv'}{v^2}$. It’s $0$ only when the numerator $u'v - uv'$ is $0$ (and $v \ne 0$). So simplify the numerator and set just that equal to $0$. Points where the denominator is $0$ are usually outside the domain anyway.

Justifying on an open domain

Many of these problems have domains like $x \gt 0$ or $t \ge 0$ with no right endpoint, so the candidates test doesn’t apply directly. Use one of these instead:

• First derivative test on the whole domain: if $f' \gt 0$ to the left of the only critical point and $f' \lt 0$ to the right of it, the critical point is the absolute maximum (and the reverse for a minimum).
• Second derivative test with one critical point: if there’s only one critical point in the domain and $f'' \lt 0$ there, it’s the absolute maximum ($f'' \gt 0$ gives the absolute minimum).
• Closed domain: if the context gives limits (a speed limit, a maximum production), use the candidates test with the endpoints.

Average cost and average rate

Two kinds of objective function come up again and again:

• Average cost per item. If $C(x)$ is the total cost of making $x$ items, the average cost per item is $A(x) = \dfrac{C(x)}{x}$. Dividing a polynomial by $x$ gives a rational function.
• Average rate. If an amount $N(t)$ is collected in time $t$, plus some “dead time” $c$ during which nothing is collected (travel, set-up), the average rate is $R(t) = \dfrac{N(t)}{t + c}$. Staying longer collects more in total, but more slowly, so there’s a best time.

Worked examples

Example 1: Minimizing average cost

A small company makes wooden hockey sticks. The total cost of making $x$ sticks in a week is

$$C(x) = 2000 + 15x + 0.05x^2 \text{ dollars}$$

How many sticks per week give the lowest average cost per stick? What is that average cost?

Solution. The average cost per stick is

$$A(x) = \frac{C(x)}{x} = \frac{2000}{x} + 15 + 0.05x, \qquad x \gt 0$$

Critical points: write $\frac{2000}{x}$ as $2000x^{-1}$:

$$A'(x) = -\frac{2000}{x^2} + 0.05 = 0 \quad\Rightarrow\quad x^2 = \frac{2000}{0.05} = 40\,000 \quad\Rightarrow\quad x = 200$$

(The negative root isn’t in the domain.)

Justify: $A''(x) = \dfrac{4000}{x^3} \gt 0$ for all $x \gt 0$, so $x = 200$ is a local minimum. It’s the only critical point in the domain, so it’s the absolute minimum.

$$A(200) = \frac{2000}{200} + 15 + 0.05(200) = 10 + 15 + 10 = 35$$

Making $200$ sticks per week gives the lowest average cost, $35 per stick.

Why is there a best value? For small $x$, the fixed cost of $2000 is shared by only a few sticks. For large $x$, the $0.05x^2$ term (overtime, wear on machines) takes over. The minimum is the balance between the two.

Example 2: Revenue with an exponential demand model

A ski hill finds that when a day pass costs $p$ dollars, the number of passes sold on a typical Saturday is about

$$n(p) = 5000e^{-0.02p}$$

What price maximizes the revenue, and what is the maximum revenue?

Solution. Revenue is price times the number sold:

$$R(p) = p \cdot n(p) = 5000pe^{-0.02p}, \qquad p \ge 0$$

Critical points: product rule, then factor out the exponential:

$$R'(p) = 5000e^{-0.02p} + 5000p\big(-0.02e^{-0.02p}\big) = 5000e^{-0.02p}(1 - 0.02p)$$

$e^{-0.02p} \gt 0$, so $R'(p) = 0$ only when $1 - 0.02p = 0$, that is, $p = 50$.

Justify: the sign of $R'(p)$ is the sign of $1 - 0.02p$: positive for $0 \le p \lt 50$ and negative for $p \gt 50$. $R$ increases and then decreases, so $p = 50$ gives the absolute maximum.

$$R(50) = 5000(50)e^{-1} = \frac{250\,000}{e} \approx 91\,969.86$$

At $50 a pass, the hill sells about $\dfrac{5000}{e} \approx 1839$ passes, for a maximum revenue of about $91 970.

Check: at $40 the revenue is about $89 866, and at $60 it’s about $90 358. Both are less. ✓

Example 3: Drug concentration

The concentration of a medication in a patient’s blood $t$ hours after a dose is

$$C(t) = 12te^{-0.4t} \text{ mg/L}, \qquad t \ge 0$$

• (a) When is the concentration highest, and what is the maximum concentration?
• (b) When is the concentration decreasing most quickly?

Solution.

(a) Product rule:

$$C'(t) = 12e^{-0.4t} + 12t\big(-0.4e^{-0.4t}\big) = 12e^{-0.4t}(1 - 0.4t)$$

$C'(t) = 0$ when $1 - 0.4t = 0$, so $t = 2.5$. For $0 \le t \lt 2.5$, $C'(t) \gt 0$; for $t \gt 2.5$, $C'(t) \lt 0$. So the concentration rises, then falls, and $t = 2.5$ gives the absolute maximum:

$$C(2.5) = 12(2.5)e^{-1} = \frac{30}{e} \approx 11.036 \text{ mg/L}$$

The concentration is highest $2.5$ hours after the dose, at about $11.036$ mg/L.

(b) “Decreasing most quickly” means $C'(t)$ is at its most negative, so we need the minimum of $C'(t)$. Find where $C''(t) = 0$:

$$\begin{aligned} C''(t) &= 12\big(-0.4e^{-0.4t}\big)(1 - 0.4t) + 12e^{-0.4t}(-0.4) \\ &= 12e^{-0.4t}\big(-0.4 + 0.16t - 0.4\big) \\ &= 12e^{-0.4t}(0.16t - 0.8) \end{aligned}$$

$C''(t) = 0$ when $t = 5$. $C''$ changes from negative to positive there, so $C'$ has its minimum at $t = 5$ (and the graph of $C$ has a point of inflection there).

$$C'(5) = 12e^{-2}(1 - 2) = -12e^{-2} \approx -1.624$$

Five hours after the dose, the concentration is dropping fastest, at about $1.624$ mg/L per hour. After that, the drop slows down as the curve levels off.

$C(t) = 12te^{-0.4t}$ peaks at $t = 2.5$ h and falls fastest at the inflection point, $t = 5$ h.

Example 4: Maximizing an average rate

A bee visits a patch of flowers. In $t$ minutes in the patch, it collects

$$N(t) = \frac{40t}{t + 9} \text{ mg of nectar}$$

Flying to the next patch takes $3$ minutes, and the bee collects nothing on the way. How long should the bee stay in each patch to maximize its average rate of collection, counting the flying time?

Solution. Each cycle takes $t + 3$ minutes (staying plus flying) and gives $N(t)$ mg. The average rate is

$$R(t) = \frac{N(t)}{t + 3} = \frac{40t}{(t + 9)(t + 3)} = \frac{40t}{t^2 + 12t + 27}, \qquad t \gt 0$$

Critical points: quotient rule. Only the numerator matters for $R'(t) = 0$:

$$\begin{aligned} R'(t) &= \frac{40(t^2 + 12t + 27) - 40t(2t + 12)}{(t^2 + 12t + 27)^2} \\ &= \frac{40(t^2 + 12t + 27 - 2t^2 - 12t)}{(t^2 + 12t + 27)^2} \\ &= \frac{40(27 - t^2)}{(t^2 + 12t + 27)^2} \end{aligned}$$

$R'(t) = 0$ when $t^2 = 27$, so $t = \sqrt{27} = 3\sqrt{3} \approx 5.196$ (the negative root isn’t in the domain).

Justify: the denominator is positive, so $R'(t)$ has the sign of $27 - t^2$: positive for $0 \lt t \lt 3\sqrt{3}$ and negative for $t \gt 3\sqrt{3}$. So $t = 3\sqrt{3}$ gives the absolute maximum.

$$R(3\sqrt{3}) = \frac{40 - 20\sqrt{3}}{3} \approx 1.786 \text{ mg/min}$$

The bee should stay about $5.2$ minutes in each patch, for an average of about $1.786$ mg of nectar per minute.

A picture of the answer. The average rate $\dfrac{N(t)}{t + 3}$ is the slope of the line from $(-3, 0)$ to the point $(t, N(t))$ on the curve. The best choice of $t$ makes that line as steep as possible, which happens when it just touches the curve: it’s tangent there. Staying longer (say $15$ minutes) gives more nectar per patch, but a less steep line, so a lower average rate.

The average rate is the slope of a line from $(-3, 0)$ to the curve. The steepest such line is tangent to the curve, at $t = 3\sqrt{3} \approx 5.2$ min.

Common mistakes

Setting the exponential factor equal to zero. In $5000e^{-0.02p}(1 - 0.02p) = 0$, the equation $e^{-0.02p} = 0$ has no solution. Don’t divide by it carelessly or try to solve it; just say ”$e^{-0.02p} \gt 0$, so $1 - 0.02p = 0$”.

Optimizing total cost instead of average cost. “Lowest cost per item” means minimize $\dfrac{C(x)}{x}$, not $C(x)$. The total cost $C(x)$ in Example 1 is smallest at $x = 0$, which isn’t a useful answer.

Forgetting the dead time in an average rate. In Example 4, dividing by $t$ instead of $t + 3$ gives $\dfrac{40}{t + 9}$, which is largest as $t \to 0$: a nonsense answer that says the bee should leave immediately. The travel time is what makes staying longer worthwhile.

Setting the whole quotient equal to zero, denominator and all. For a rational derivative, set only the numerator equal to $0$. Then check that the denominator isn’t $0$ at your answer and that it’s in the domain.

No justification on an open domain. With a domain like $t \gt 0$, there are no endpoints to compare. Give a sign argument for $f'$ on both sides of the critical point, or use $f''$ and note it’s the only critical point.

Answering with the wrong quantity. If the question asks for the price that maximizes revenue, the answer is $50, not $91 970. Read the question again before you write your final sentence, and include units.

Practice

1. (Warm-up) The total cost of producing $x$ units of a product is $C(x) = 500 + 4x + 0.05x^2$ dollars. Find the production level that minimizes the average cost per unit, and the minimum average cost.

Solution

$$A(x) = \frac{C(x)}{x} = \frac{500}{x} + 4 + 0.05x, \qquad x \gt 0$$

$A'(x) = -\dfrac{500}{x^2} + 0.05 = 0$ gives $x^2 = 10\,000$, so $x = 100$.

$A''(x) = \dfrac{1000}{x^3} \gt 0$, and $x = 100$ is the only critical point, so it’s the absolute minimum.

$A(100) = 5 + 4 + 5 = 14$. Producing $100$ units gives the lowest average cost, $14 per unit.

2. (Warm-up) Find the maximum value of $f(x) = 5xe^{-x}$ for $x \ge 0$.

Solution

$f'(x) = 5e^{-x} - 5xe^{-x} = 5e^{-x}(1 - x)$. Since $e^{-x} \gt 0$, $f'(x) = 0$ only at $x = 1$.

$f'(x) \gt 0$ for $0 \le x \lt 1$ and $f'(x) \lt 0$ for $x \gt 1$, so $x = 1$ gives the absolute maximum: $f(1) = \dfrac{5}{e} \approx 1.839$.

3. (Core) A trucking company sends a truck on a $500$ km trip. The driver is paid $30 per hour, and fuel costs $0.004v^2$ dollars per hour when the truck travels at $v$ km/h. The truck must travel between $60$ km/h and $100$ km/h. What speed minimizes the total cost of the trip? Round to 1 decimal place.

Solution

The trip takes $\dfrac{500}{v}$ hours, and the cost per hour is $30 + 0.004v^2$:

$$C(v) = \left(30 + 0.004v^2\right)\frac{500}{v} = \frac{15\,000}{v} + 2v, \qquad 60 \le v \le 100$$

$C'(v) = -\dfrac{15\,000}{v^2} + 2 = 0$ gives $v^2 = 7500$, so $v = 50\sqrt{3} \approx 86.6$, which is in the domain.

Candidates:

$v$ (km/h)	$60$	$50\sqrt{3} \approx 86.6$	$100$
$C(v)$ ($)	$250 + 120 = 370$	$\approx 346.41$	$150 + 200 = 350$

The cheapest speed is about $86.6$ km/h, for a total cost of about $346.41.

4. (Core) A science centre finds that when the admission price is $p$ dollars, the daily attendance is about $n(p) = 2400(1.08)^{-p}$. What price maximizes the daily revenue? Find the maximum revenue to the nearest dollar.

Solution

$$R(p) = 2400p(1.08)^{-p}, \qquad p \ge 0$$

Product rule, using $\dfrac{d}{dp}\big[(1.08)^{-p}\big] = -(1.08)^{-p}\ln 1.08$:

$$R'(p) = 2400(1.08)^{-p} - 2400p(1.08)^{-p}\ln 1.08 = 2400(1.08)^{-p}\big(1 - p\ln 1.08\big)$$

$(1.08)^{-p} \gt 0$, so $R'(p) = 0$ when $p = \dfrac{1}{\ln 1.08} \approx 12.99$.

$R'(p) \gt 0$ for smaller $p$ and $R'(p) \lt 0$ for larger $p$, so this is the absolute maximum. The best price is about $12.99.

At that price, $(1.08)^{-p} = e^{-p\ln 1.08} = e^{-1}$, so the attendance is $\dfrac{2400}{e} \approx 883$ people and

$$R \approx 12.9936 \times \frac{2400}{e} \approx 11\,472$$

The maximum daily revenue is about $11 472.

5. (Core) The concentration of a drug in the bloodstream $t$ hours after an injection is $C(t) = \dfrac{5t}{t^2 + 4}$ mg/L. When is the concentration greatest, and what is the maximum concentration?

Solution

$$C'(t) = \frac{5(t^2 + 4) - 5t(2t)}{(t^2 + 4)^2} = \frac{5(4 - t^2)}{(t^2 + 4)^2}, \qquad t \ge 0$$

$C'(t) = 0$ when $t = 2$ (the root $t = -2$ is outside the domain). $C'(t) \gt 0$ for $0 \le t \lt 2$ and $C'(t) \lt 0$ for $t \gt 2$, so this is the absolute maximum.

$C(2) = \dfrac{10}{8} = 1.25$. The concentration is greatest after $2$ hours, at $1.25$ mg/L.

6. (Core) A printed poster must have $384$ cm² of printed area, with margins of $4$ cm at the top and bottom and $2$ cm on each side. What dimensions of paper use the least area? Round to 2 decimal places.

Solution

Let the printed region be $x$ cm wide and $\dfrac{384}{x}$ cm tall. The paper is $x + 4$ cm wide and $\dfrac{384}{x} + 8$ cm tall:

$$P(x) = (x + 4)\left(\frac{384}{x} + 8\right) = 384 + 8x + \frac{1536}{x} + 32 = 416 + 8x + \frac{1536}{x}, \qquad x \gt 0$$

$P'(x) = 8 - \dfrac{1536}{x^2} = 0$ gives $x^2 = 192$, so $x = 8\sqrt{3} \approx 13.86$.

$P''(x) = \dfrac{3072}{x^3} \gt 0$, and it’s the only critical point, so it’s the absolute minimum.

The printed region is $8\sqrt{3}$ cm wide and $\dfrac{384}{8\sqrt{3}} = 16\sqrt{3}$ cm tall. The paper is $8\sqrt{3} + 4 \approx 17.86$ cm wide by $16\sqrt{3} + 8 \approx 35.71$ cm tall.

7. (Core) A company’s total cost of producing $x$ batches of a chemical is $C(x) = 100e^{0.02x}$ dollars. Find the number of batches that minimizes the average cost per batch, and the minimum average cost to the nearest cent.

Solution

$$A(x) = \frac{100e^{0.02x}}{x}, \qquad x \gt 0$$

Quotient rule:

$$A'(x) = \frac{100(0.02)e^{0.02x} \cdot x - 100e^{0.02x}}{x^2} = \frac{100e^{0.02x}(0.02x - 1)}{x^2}$$

$A'(x) = 0$ when $0.02x - 1 = 0$, so $x = 50$. $A'(x) \lt 0$ for $0 \lt x \lt 50$ and $A'(x) \gt 0$ for $x \gt 50$, so this is the absolute minimum.

$A(50) = \dfrac{100e}{50} = 2e \approx 5.44$. Making $50$ batches gives the lowest average cost, about $5.44 per batch.

8. (Challenge) A forager collects $N(t) = \dfrac{at}{t + b}$ units of food in $t$ minutes at a patch, where $a$ and $b$ are positive constants. Travelling between patches takes $c$ minutes. Show that the average rate of collection, including travel time, is greatest when $t = \sqrt{bc}$. Check your formula against Example 4.

Solution

$$R(t) = \frac{at}{(t + b)(t + c)} = \frac{at}{t^2 + (b + c)t + bc}, \qquad t \gt 0$$

The numerator of $R'(t)$ is

$$a\big(t^2 + (b + c)t + bc\big) - at\big(2t + b + c\big) = a\big(bc - t^2\big)$$

The denominator of $R'(t)$ is a square, so it’s positive. So $R'(t) = 0$ when $t^2 = bc$, giving $t = \sqrt{bc}$, and $R'(t)$ changes from positive to negative there (since $bc - t^2$ does). That makes $t = \sqrt{bc}$ the absolute maximum.

In Example 4, $b = 9$ and $c = 3$, so $t = \sqrt{27} = 3\sqrt{3}$. ✓ (Notice that $a$ doesn’t matter: a richer patch doesn’t change how long to stay.)

9. (Challenge) A salmon swims $L$ km upstream against a current of $4$ km/h. Its speed relative to the water is $v$ km/h, with $v \gt 4$. The energy it uses per hour is proportional to $v^3$, so the total energy for the trip is

$$E(v) = k v^3 \cdot \frac{L}{v - 4}$$

where $k$ and $L$ are positive constants. Find the speed that minimizes the energy used.

Solution

$kL$ is a positive constant, so it’s enough to minimize $f(v) = \dfrac{v^3}{v - 4}$ for $v \gt 4$.

$$f'(v) = \frac{3v^2(v - 4) - v^3}{(v - 4)^2} = \frac{2v^3 - 12v^2}{(v - 4)^2} = \frac{2v^2(v - 6)}{(v - 4)^2}$$

For $v \gt 4$, $f'(v) = 0$ only at $v = 6$. $f'(v) \lt 0$ for $4 \lt v \lt 6$ and $f'(v) \gt 0$ for $v \gt 6$, so $v = 6$ gives the absolute minimum.

The salmon uses the least energy swimming at $6$ km/h relative to the water, which is $1.5$ times the speed of the current. (Its speed relative to the riverbank is then $6 - 4 = 2$ km/h.)