B. Derivatives and Their Applications · Applications of Derivatives: Motion, Rates and Optimization

Modelling with Derivatives: The Foraging Bird and Other Real Problems

B2.4 B2.5About 75 minutes

Learning goals

I can follow the modelling cycle: understand, model, solve, interpret, validate and communicate.
I can solve an optimization model numerically, graphically and algebraically.
I can explain the marginal value rule for the foraging bird.
I can build a model from data, use derivatives to make a recommendation, and state its assumptions and limits.

Words to know

mathematical model
A function or equation that represents a real situation, built on stated assumptions.

assumption
A simplification accepted in order to build a model.

validate
Check a model's predictions against data or common sense.

marginal value theorem
In foraging: leave a patch when the instantaneous rate of gain falls to the best average rate, E′(t) = E(t)/(t + T).

extrapolation
Using a model outside the range of the data it was built from; often unreliable.

Thinking like a modeller

The last expectation in this strand, B2.5, asks you to apply a mathematical model and the concepts of the derivative to solve a real problem, and to interpret and communicate the results. This lesson is about that whole process, not just the calculus in the middle. The curriculum's sample is a lovely one from ecology: how long should a foraging bird stay in a patch of food before flying to the next one?

Every model starts with assumptions, and every answer should end with a check against reality and a clear explanation of what the model can and cannot say.

The modelling cycle. The calculus (step 3) is only one part. Interpreting, validating and communicating matter just as much, and a failed check sends you back to improve the model.

The foraging bird

A chickadee feeds on seeds in patches (shrubs, feeders) spread through a forest. Inside a patch, it finds food quickly at first, then more slowly as the easy seeds run out: diminishing returns. Getting to the next patch costs travel time T, during which it gains nothing.

Assumptions: every patch is alike; E(t) is the energy gained after t minutes in a patch; E increases but more and more slowly; the travel time T between patches is the same each time; the bird wants to maximize its average rate of energy gain over a whole cycle (travel + feeding).

Model: each cycle takes t + T minutes and gains E(t), so the average rate is

R(t) = E(t) ÷ (t + T).

Too short a stay wastes the long trip. Too long a stay means feeding in a nearly empty patch. The best t is in between.

Worked example

Example 1: Solving the bird's problem three ways

Energy gained in a patch is E(t) = 60t/(t + 2) joules after t minutes, and the travel time is T = 2 min. How long should the bird stay?

Numerically: R(t) = 60t/(t + 2)². R(1) = 60/9 ≈ 6.67, R(2) = 120/16 = 7.5, R(3) = 180/25 = 7.2 J/min. The best is near t = 2.
Algebraically: write R(t) = 60t(t + 2)⁻². Product rule: R′(t) = 60(t + 2)⁻² − 120t(t + 2)⁻³ = 60(t + 2)⁻³[(t + 2) − 2t] = 60(2 − t)/(t + 2)³.
R′(t) = 0 at t = 2; R′ > 0 for t < 2 and R′ < 0 for t > 2, so t = 2 gives the absolute maximum for t > 0. R(2) = 7.5 J/min.
Graphically: the average rate E(t)/(t + T) is the slope of the line from (−T, 0) to the point (t, E(t)) on the energy curve. The steepest such line is the one that is tangent to the curve. Draw it from (−2, 0): it touches at t = 2.

Stay 2 minutes in each patch, for an average gain of 7.5 J/min.

The graphical strategy. Each line from (−2, 0) to a point on the energy curve has slope equal to the average rate for that leaving time. The steepest possible line is tangent to the curve, at t = 2. Staying until t = 6 gives a shallower line, so a lower average rate.

The marginal value rule

Setting R′(t) = 0 for R = E(t)(t + T)⁻¹ gives E′(t)(t + T)⁻¹ − E(t)(t + T)⁻² = 0, so

E′(t) = E(t) ÷ (t + T).

In words: leave the patch when the instantaneous rate of gain drops to the best average rate. Ecologists call this the marginal value theorem. Notice it is the same idea as marginal revenue = marginal cost: stop when the next minute is worth less than average.

Worked example

Example 2: When algebra cannot finish the job

Suppose instead E(t) = 100(1 − e^(−0.5t)) joules and T = 2 min. Find the best time to leave.

Rule: E′(t) = E(t)/(t + T). E′(t) = 50e^(−0.5t), so 50e^(−0.5t)(t + 2) = 100(1 − e^(−0.5t)).
Divide by 100 and rearrange: e^(−0.5t)(0.5t + 1) = 1 − e^(−0.5t), so e^(−0.5t)(0.5t + 2) = 1.
This cannot be solved with algebra alone (t is both inside and outside the exponential). Use technology or a table: try t = 2: e^(−1)(3) ≈ 1.104 (too big); t = 2.5: e^(−1.25)(3.25) ≈ 0.931 (too small); t = 2.3: e^(−1.15)(3.15) ≈ 0.997. So t ≈ 2.29.
R(2.29) = 100(1 − e^(−1.146)) ÷ 4.29 ≈ 68.2 ÷ 4.29 ≈ 15.9 J/min.

Leave after about 2.3 minutes, for about 15.9 J/min. Using numerical, graphical and algebraic strategies together is exactly what B2.5 asks for.

Worked example

Example 3: Modelling from data and communicating the result

A kayak rental shop on a Muskoka lake tried five daily prices over one summer and recorded the average number of rentals: $30: 54, $35: 48, $40: 43, $45: 37, $50: 32. Build a model, find the price that maximizes revenue, check it, and write a short conclusion.

Model: the points are close to a line. Use the first and last points: slope = (32 − 54)/(50 − 30) = −1.1, so N(p) = 54 − 1.1(p − 30) = 87 − 1.1p. (A regression line on a graphing calculator gives almost the same result.)
Solve: R(p) = p(87 − 1.1p) = 87p − 1.1p². R′(p) = 87 − 2.2p = 0 at p ≈ 39.55. R″ = −2.2 < 0, a maximum. R(39.55) ≈ $1720.
Validate: from the data, revenues were 30 × 54 = $1620, $1680, $1720, $1665 and $1600. The best observed price ($40, $1720) agrees with the model.
Communicate: "Based on one summer of data, daily revenue is highest at a price of about $40 (about 43 rentals, $1720 a day). The model assumes demand changes linearly with price and that weather and season stay similar. It should not be trusted far outside $30 to $50. Costs were not included, so this maximizes revenue, not profit."

Charge about $40 a day. The conclusion states the answer, the evidence, the assumptions and the limits.

Left: the kayak data with the linear demand model. Right: the revenue model R(p) = p(87 − 1.1p) with the observed revenues (points). The model's maximum is near $39.55, and the best observed price was $40.

Just for fun

The marginal value theorem was first written down for birds and bees, but it also describes students at a buffet: the first plate is fast and rewarding, the fifth is slow going, and at some point the dessert table across the room is the better "patch".

Hands-on activity

The cereal-patch foraging game

You need: 3 bowls, 3 cups of dry rice (or dry beans), 3 small handfuls of round cereal pieces, a spoon, a phone stopwatch, and a helper. (Keep small pieces away from young children.)

Mix one handful of cereal into the rice in each bowl: each bowl is a "patch". Put the bowls in different rooms; walking between rooms is your "travel time" T. Time one trip between rooms.
Using only the spoon and one hand, pick cereal pieces out of a bowl for 2 minutes while your helper records your total count every 15 seconds.
Plot total pieces E against time t. You should see diminishing returns.
On your graph, mark the point (−T, 0) and draw the line from it that just touches your curve. Read off the best leaving time.
Test it: play again, leaving each bowl at your best time, and compare your total pieces per minute (including travel) with a round where you always stay 2 minutes.

What to look for: the curve rises fast and then flattens as the easy pieces run out. With a long trip between rooms, the tangent line touches later, so you should stay longer; with a short trip, leave sooner. Your best-time rounds should give a higher overall rate, though the difference can be small.

Your turn

For E(t) = 60t/(t + 2) with travel time T = 8 min, use the rule E′(t) = E(t)/(t + T) to find the best stay.

Compare your answer to Question 1 with Example 1 (T = 2). How does a longer trip change the best stay? Why does this make sense?

Using R(t) = 60t/(t + 2)², complete the table: R(1.5), R(2), R(2.5). Does it confirm t = 2?

Using the kayak model N(p) = 87 − 1.1p, predict the number of rentals at $42.

Each kayak rental costs the shop $8 in cleaning and wear. Find the price that maximizes daily profit.

Name three assumptions in the foraging-bird model.

For E(t) = 100(1 − e^(−0.5t)) and T = 4, the best stay satisfies e^(−0.5t)(0.5t + 3) = 1. Use a table to estimate t to one decimal place.

More practice: warm-up

In the modelling cycle, what step comes after solving?

In R(t) = E(t)/(t + T), what does t + T represent?

What does "diminishing returns" mean for E(t)?

For E(t) = 60t/(t + 2), find E(2).

Why is no model perfect?

More practice: on your own

For E(t) = 60t/(t + 1) and T = 9, find the best stay and the best average rate.

A fishing boat's catch in one fishing ground is C(t) = 200(1 − e^(−0.2t)) kg after t hours, and moving to a new ground takes 3 h. The best stay satisfies e^(−0.2t)(0.2t + 1.6) = 1. Estimate it.

The kayak shop later tries $55 and gets 30 rentals. The model predicts 87 − 60.5 = 26.5. What does this suggest?

Explain in words why the graphical strategy (tangent from (−T, 0)) gives the best leaving time.

Write a two-sentence conclusion for Example 1 that a non-mathematician (say, a birdwatcher) would understand.

Challenge

Show that for E(t) = Mt/(t + a) (M, a > 0) and travel time T, the best stay is t = √(aT).

Prove the marginal value rule: if R(t) = E(t)/(t + T) has a maximum at t₀ > 0, then E′(t₀) = E(t₀)/(t₀ + T). Explain it as a statement about a tangent line.

For E(t) = 100(1 − e^(−0.5t)), the best stays were about 2.29 min for T = 2 and 3.01 min for T = 4. Explain, using the tangent picture, why the best stay always increases when T increases (for any concave-down E).

Check your understanding

Quiz

In the foraging model, what does R(t) = E(t)/(t + T) measure?

The marginal value rule says the bird should leave the patch when…

For E(t) = 60t/(t + 2) and T = 2, what is the best stay?

If the travel time between patches increases, the best time to stay in each patch…

Which belongs in a good written conclusion to a modelling problem?

An equation such as e^(−0.5t)(0.5t + 2) = 1 cannot be solved by algebra alone. What is a good strategy?

For parents and teachers

Big idea (B2.5): a model is only useful if its answer is interpreted, checked against reality and clearly communicated. The foraging bird (the curriculum's own sample) shows how one optimization can be solved with a table, a graph and algebra, and all three should agree.

How to help: after your teen solves a problem, ask them to explain the answer to you as if you were the shop owner or the birdwatcher: no symbols, just the recommendation, the reason and one limitation. The cereal-patch game is a fun way to feel diminishing returns.

Common difficulties:

Stopping at the number without interpreting it.
Not stating assumptions, or not noticing when a prediction is far outside the data.
Trying to force an algebraic solution when a table or graph is the right tool.
Product-rule errors when differentiating E(t)(t + T)⁻¹.