Optimization, question 16

Topic: Optimization
Difficulty: Challenging
Suggested time: 7 min
Marks: 4
Theory: UBC Math 100 §8 CLP-1 §3.5Key results

A bird feeds in patches of berries. After $t$ minutes in a patch it has gained $f(t)$ units of energy, given below, so the patch gives up food more and more slowly. Flying to the next patch takes 6 minutes, with no food on the way. How long should the bird stay in each patch to make its long-run rate of energy gain, energy per minute of feeding and flying together, as large as possible?

f(t) = \frac{60t}{t + 3}

Give the time in minutes.

Your answer

min

An exact expression, or a decimal to two decimal places.

How sure are you?

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Theory UBC Math 100 §8 CLP-1 §3.5Key results

Worked solution

Strategy

The bird should maximise energy per minute of the whole cycle, travel included. Write that rate as a function of the time spent in a patch, differentiate with the quotient rule, and check the sign of R′ on either side.

Working

$R(t) = \frac{f(t)}{t + 6} = \frac{60t}{t^{2} + 9t + 18}, \qquad t > 0$
Energy from one patch over the length of one cycle.
$R'(t) = \frac{60(t^{2} + 9t + 18) - 60t(2t + 9)}{(t^{2} + 9t + 18)^{2}} = \frac{60(18 - t^{2})}{(t^{2} + 9t + 18)^{2}}$
Quotient rule; the 9t terms cancel in the numerator.
$R'(t) = 0 \iff t^{2} = 18 \iff t = \sqrt{18} = 3\sqrt{2}$
Only the positive root is a time.
$R'(t) > 0 \text{ for } 0 < t < 3\sqrt{2}, \qquad R'(t) < 0 \text{ for } t > 3\sqrt{2}$
The rate rises and then falls, so this is the global maximum for t > 0.
$R\bigl(3\sqrt{2}\bigr) = 60 - 40\sqrt{2} \approx 3.43$
The best long-run rate, in units of energy per minute, for reference.

Answer

t = 3\sqrt{2} \approx 4.24 \text{ minutes}

Check

Staying 4 or 4.5 minutes gives a long-run rate of 24/7 ≈ 3.4286 units a minute either way, just under the 3.4315 at 3√2 ≈ 4.24.

Where this goes wrong

Leaving the 6 minutes of flight out of the rate: f(t) only rises and f(t)/t only falls, so neither has a best stay, and the travel time is what makes a longer stay worthwhile. Or stopping at t² = 18 and answering 18 instead of 3√2.

Leans on

The quotient rule, from Differentiation rules

Try a similar question

A furniture workshop’s total cost of making

q

chairs a month is

C(q)

dollars, given below. Find the smallest possible average cost per chair.

C(q) = 2560 + 25q + 0.1q^{2}

My working

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Optimization practice question 16