Lesson 4 of 4Applications of Derivatives

Unit 4 · Lesson 4 of 4

Optimization in Business and Science

The same calculus finds the ticket price that earns the most, the bus fare that maximizes revenue under an exponential demand model, the production level with the lowest cost per item, and how long a foraging bird should stay in a berry patch. Then you interpret the result, with units and a domain check.

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Key things to remember

1Revenue $R = (price) (quantity sold)$ ; profit $P (x) = R (x) - C (x)$ .
2Marginal revenue is $R^{'} (x)$ and marginal cost is $C^{'} (x)$ ; a maximum profit inside the domain occurs where $R^{'} (x) = C^{'} (x)$ .
3Average cost is $\frac{C ( x )}{x}$ ; differentiate it (it is a rational function) to find the lowest cost per item.
4For $R (x) = k x b^{- x}$ with $k > 0$ and $b > 1$ , the maximum is at $x = \frac{1}{ln b}$ ; for the bus fare, $\frac{1}{ln 1.15} \approx 7.16$ .
5Compare numeric (table), graphical and algebraic (derivative) strategies: the algebraic one gives the exact value.
6Always answer in context with units, a domain check and sensible rounding.

Set a concert ticket at 0 dollars and the hall fills, but you earn nothing. Set it at 500 dollars and you earn nothing either, because nobody comes. Somewhere in between is the price that brings in the most money. Businesses face trade-offs like this constantly, and so do living things: a bird that stays too long in one berry patch wastes time searching for the last hidden berries, but leaving costs time finding a new patch. In both cases you build a model, differentiate, and find where the derivative is zero, just as with boxes and fences. The new part is that the models now include rational and exponential functions, and the answer has to make sense in the real situation.

Business vocabulary: if $x$ items sell at a price of $p$ dollars each, the REVENUE is $R = x p$ . A DEMAND function gives how many items sell at each price (or the price at which a given number will sell). If $C (x)$ is the total cost of making $x$ items, the PROFIT is $P (x) = R (x) - C (x)$ . The derivatives $R^{'} (x)$ and $C^{'} (x)$ are called MARGINAL revenue and MARGINAL cost: roughly, the extra revenue or cost from one more item. Inside the domain, profit can be greatest only where $P^{'} (x) = 0$ , that is, where marginal revenue equals marginal cost (then check that it is a maximum, and compare any endpoints).

Fig 1Revenue against ticket price for the linear demand q = 1200 − 40p. Top: R(p) = p(1200 − 40p) is a downward parabola with a horizontal tangent at its peak, p = 15. Bottom: R′(p) = 1200 − 80p is positive before p = 15, zero there (marginal revenue = 0) and negative after.

VideoSal Khan uses derivatives to find the production level that maximizes a shoe factory's profit, the same revenue-minus-cost reasoning as this lesson.Watch on YouTube

Revenue with a linear demand

A theatre finds that at a price of $p$ dollars it sells $q = 1200 - 40 p$ tickets. Revenue is $R (p) = p (1200 - 40 p) = 1200 p - 40 p^{2}$ on the closed interval $0 \leq p \leq 30$ (above 30 dollars, no tickets sell). Then $R^{'} (p) = 1200 - 80 p = 0$ at $p = 15$ . Comparing $R (0) = 0$ , $R (15) = 15 (600) = 9000$ and $R (30) = 0$ , the best price is 15 dollars: 600 tickets and 9000 dollars of revenue. Raising the price above 15 dollars loses more in ticket sales than it gains per ticket.

Exponential demand: the bus fare

Suppose the number of bus riders per day from the suburbs to downtown is $1200 (1.15)^{- x}$ when the fare is $x$ dollars: each extra dollar loses about 13% of the riders, since $(1.15)^{- 1} \approx 0.87$ . Revenue is fare times riders, $R (x) = 1200 x (1.15)^{- x}$ . The product rule, with $\frac{d}{d x} (1.15)^{- x} = - (1.15)^{- x} ln 1.15,$ gives

R^{'} (x) = 1200 (1.15)^{- x} - 1200 x (1.15)^{- x} ln 1.15 = 1200 (1.15)^{- x} (1 - x ln 1.15)

An exponential is never zero, so R′(x) = 0 only when 1 − x ln 1.15 = 0, that is, x = 1/ln 1.15 ≈ 7.16.

Fare x (dollars)	Riders per day	Revenue R(x) (dollars per day)
$5$	$596.6$	$2983.06$
$6$	$518.8$	$3112.76$
$7$	$451.1$	$3157.87$
$7.16$	$441.1$	$3158.62$
$8$	$392.3$	$3138.26$
$9$	$341.1$	$3070.03$

Numerical check of the bus-fare model: revenue climbs to a peak near a fare of 7.16 dollars and then falls. Riders are shown to one decimal because the model gives an average.

Modelling: the foraging bird

A bird foraging for berries gets a net $E (t) = \frac{3000 t}{t + 4}$ joules of food energy by spending $t$ minutes in a patch. The returns shrink as it stays, because the easy berries go first. Finding a new patch takes $2$ minutes on average (using negligible energy). The bird should not maximize $E$ itself, which keeps growing toward $3000$ J; it should maximize its AVERAGE RATE of energy gain over the whole cycle of flying and foraging:

A (t) = \frac{E ( t )}{t + 2} = \frac{3000 t}{( t + 4 ) ( t + 2 )} = \frac{3000 t}{t ^{2} + 6 t + 8}, t > 0

Average rate of energy gain, in joules per minute, for a bird that stays t minutes in each patch.

Numerically: a table of $A (t)$ gives $200$ J/min at $t = 1$ , $250$ at $t = 2$ , about $257.35$ at $t = 2.8$ , about $257.36$ at $t = 2.83$ , about $257.14$ at $t = 3$ and $250$ at $t = 4$ . The peak is near $2.8$ min, but a table cannot pin it down exactly.
Graphically: $A (t)$ is the slope of the line from the point $(- 2, 0)$ (the moment the bird starts searching) to the point $(t, E (t))$ on the energy curve. That slope is steepest when the line just touches the curve, which is where $E^{'} (t) = \frac{E ( t )}{t + 2}$ . Graphing $A (t)$ with technology shows the same peak.
Algebraically: the quotient rule gives $A^{'} (t) = \frac{3000 ( 8 - t ^{2} )}{( t ^{2} + 6 t + 8 ) ^{2}},$ which is zero at $t = 8 = 22 \approx 2.83$ min. It is exact and explains WHY the answer is what it is.

Loading interactive…

Choose Ticket revenue. Change how many tickets would go at a price of 0 dollars and how many sales are lost for each 1-dollar increase, drag the price, and watch the revenue graph and the sign of R′; then press Find the optimum to see the price where R′ = 0 and the endpoint check.

Interpret and communicate the result

State the domain the context allows and check its endpoints. Round to a sensible real-world value (a fare to the cent, a number of items to a whole number) and check that rounding does not change the conclusion. Finish with a sentence that answers the question, with units: "A fare of about 7.16 dollars gives the maximum daily revenue, about 3159 dollars, from about 441 riders."

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

A theatre charges 12 dollars a ticket and sells 400 tickets. For each 1-dollar increase in price, 20 fewer tickets sell. What price maximizes revenue?

Example 2

Bus riders per day are

1200 (1.15)^{- x}

at a fare of

x

dollars. What fare maximizes revenue? Solve exactly and verify.

Example 3

Making

x

items costs

C (x) = 10000 + 20 x + 0.01 x^{2}

dollars. How many items give the lowest average cost per item, and what is that cost?

Example 4

A bird gets

E (t) = \frac{3000 t}{t + 4}

joules in

t

minutes in a patch, and takes

2

minutes to find each patch. How long should it stay to maximize its average rate of energy gain?

Where marks get lost

Maximizing the wrong quantity: the number of riders or the energy $E (t)$ instead of revenue or the AVERAGE rate $\frac{E ( t )}{t + 2}$ . Reread what the question wants as large as possible.
Differentiating $(1.15)^{- x}$ as $- x (1.15)^{- x - 1}$ (the power rule). The variable is in the exponent, so the derivative is $- (1.15)^{- x} ln 1.15$ .
Setting the whole product $1200 (1.15)^{- x} (1 - x ln 1.15)$ equal to zero and trying to solve $1200 (1.15)^{- x} = 0$ . An exponential is never zero; only the bracket can be.
Leaving the answer as $x \approx 7.155$ without saying it is a fare in dollars, or rounding a number of items to a decimal. Interpret in context and round sensibly.
Ignoring the domain: prices, quantities and times cannot be negative, and on a closed domain the endpoints must be compared.

Flashcards 8

Revenue

Profit

Marginal revenue / marginal cost

Condition for maximum profit

+ 4 more

Practice

At a price of

p

dollars,

q = 1200 - 40 p

tickets sell. What price, in dollars, maximizes revenue?

dollars

A club charges 20 dollars and sells 300 memberships. Each 1-dollar increase loses 10 members. What price, in dollars, maximizes revenue?

dollars

Bus riders per day are

1200 (1.15)^{- x}

at a fare of

x

dollars. Find the fare that maximizes revenue, to the nearest cent (in dollars, e.g. 1.23).

dollars

At the revenue-maximizing fare

x = \frac{1}{ln 1.15}

, how many riders per day does the model

1200 (1.15)^{- x}

predict? Round to the nearest whole number.

riders

The cost of making

x

items is

C (x) = 4000 + 10 x + 0.1 x^{2}

dollars. How many items minimize the average cost per item?

items

The concentration of a medicine in the blood is

C (t) = 20 t e^{- 0.5 t}

mg/L,

t

hours after a dose. At what time, in hours, is the concentration greatest?

For the foraging bird,

A (t) = \frac{3000 t}{( t + 4 ) ( t + 2 )}

. Find the time in the patch, in minutes, that maximizes

A (t)

, to 2 decimal places.

min

Graphically, the bird's average rate

\frac{E ( t )}{t + 2}

is the slope of which line?

A company's profit is

P (x) = R (x) - C (x)

. At a production level that maximizes profit (not at an endpoint of the domain), which is true?

Match each condition to what it finds.

R^{'} (x) = 0

R^{'} (x) = C^{'} (x)

\frac{d}{d x} (\frac{C ( x )}{x}) = 0

\frac{d}{d t} (\frac{E ( t )}{t + 2}) = 0

What is the derivative of a revenue function,

R^{'} (x)

, called? (Two words.)

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