Differential-Equations · Unit 5 · Video 6 · Interactive Practice

Don't Solve It, Lower It: Finding the Maximum Sustainable Harvest

IKey Formulas

Formula	Name	What it tells you
$\dfrac{dy}{dt} = ay - by^2 - h$	Logistic growth with constant harvest	$h$ carries no $y$ : the same amount leaves every day, whatever the stock
$f_h(y) = f_0(y) - h$	Lowering	Same shape, same peak position; only the height changes
$h_m = f_0\!\left(\tfrac{a}{2b}\right) = \dfrac{a^2}{4b}$	Maximum sustainable harvest	The peak's own height; at $h = h_m$ the rate is $-b\left(y - \tfrac{a}{2b}\right)^2 \le 0$
$y = \dfrac{a \pm \sqrt{a^2 - 4bh}}{2b}$	Critical points	Two if $h < h_m$ , one ( $y = \tfrac{a}{2b}$ ) if $h = h_m$ , none if $h > h_m$

Key Insight: Harvesting lowers the growth curve without reshaping it. The two critical points, unstable below and stable above, slide together, merge at the peak $y = \tfrac{a}{2b}$ when $h$ reaches the peak's height $h_m = \tfrac{a^2}{4b}$ , and vanish beyond it — after which every population reaches zero in finite time.

IILowering the Parabola

Harvesting lowers the growth curve by $h$ at every population: where does that move the critical points?

Harvest h 1.5

IIIWho Survives the Harvest

At each harvest rate, which starting populations level off, and which run down to an empty pen?

Harvest Start y(0) 3

IVThe Margin Below the Maximum

Just below $h_m$ the critical points reopen: how much room to fall does a small cut in the harvest buy?

Harvest h 1.5

💡 For any $a$ and $b$ the gap is $\dfrac{\sqrt{a^2 - 4bh}}{b}$ , which is $\sqrt{1 - h/h_m}$ times the capacity $\tfrac{a}{b}$ : a cut of 1% below $h_m$ leaves a margin of 10% of the capacity.

VQuiz Questions

Problem 1 · The Maximum Sustainable Harvest

A salmon pen follows $\dfrac{dy}{dt} = 6y - 2y^2 - h$ . What is the maximum sustainable harvest $h_m$ ?

✅ Correct! The peak of

6y - 2y^2

sits at

y = \tfrac{3}{2}

, and its height there is

\tfrac{9}{2} = \tfrac{a^2}{4b}

❌ That is where the peak sits.

y = \tfrac{a}{2b} = \tfrac{3}{2}

is a population, not a harvest rate.

h_m

is the peak's height: evaluate

6y - 2y^2

y = \tfrac{3}{2}

❌ That is the carrying capacity.

\tfrac{a}{b} = 3

is the upper crossing when

h = 0

. The crossings merge when the curve is lowered by the height of its peak, not by the position of a root.

❌ The crowding term is missing.

ay = 6 \cdot \tfrac{3}{2} = 9

is only the first term at the peak; the crowding term takes away

by^2 = 2 \cdot \tfrac{9}{4} = \tfrac{9}{2}

❌ Not quite. Find the peak of

6y - 2y^2

and take its height.

Show solution

Here $a = 6$ and $b = 2$ . The parabola $6y - 2y^2$ peaks where its derivative vanishes:

6 - 4y = 0 \quad\Longrightarrow\quad y = \tfrac{3}{2} = \tfrac{a}{2b}

Its height there, in the factored form $y(a - by)$ :

\tfrac{3}{2}\left(6 - 2\cdot\tfrac{3}{2}\right) = \tfrac{3}{2}\cdot 3 = \tfrac{9}{2}

This matches $h_m = \dfrac{a^2}{4b} = \dfrac{36}{8} = \dfrac{9}{2}$ . Lowering the curve by more than $\tfrac{9}{2}$ leaves it entirely below the axis, with no critical point.

Problem 2 · A Bad Season at the Borderline

A pen is harvested at exactly its maximum rate, $\dfrac{dy}{dt} = 6y - 2y^2 - \tfrac{9}{2}$ , and the stock sits at the critical point $y = \tfrac{3}{2}$ . A bad season knocks the population down to $y = 1.4$ , and the harvest continues unchanged. What happens next?

✅ Correct! The borderline critical point is semi-stable, and its unstable side is the one facing extinction: nothing below

\tfrac{3}{2}

can catch the fall.

❌ That is what a stable point would do. Here

\dfrac{dy}{dt} = -2\left(y - \tfrac{3}{2}\right)^2 \le 0

on both sides, so below

\tfrac{3}{2}

the population decreases — away from the critical point.

❌ There is no lower critical point. The perfect square

-2\left(y - \tfrac{3}{2}\right)^2

vanishes only at

y = \tfrac{3}{2}

, so no level below it can stop the decline.

❌ With a harvest, $y = 0$ is not a critical point. At

y = 0

the rate is

-\tfrac{9}{2}

, not

0

, so the solution crosses zero at a definite time instead of creeping toward it.

❌ Not quite. Factor the right-hand side and read its sign below

y = \tfrac{3}{2}

Show solution

At the borderline the right-hand side is a perfect square:

6y - 2y^2 - \tfrac{9}{2} = -2\left(y^2 - 3y + \tfrac{9}{4}\right) = -2\left(y - \tfrac{3}{2}\right)^2 \le 0

It vanishes only at $y = \tfrac{3}{2}$ , so starting from $y = 1.4$ the population decreases with no critical point below to settle on. It is semi-stable: stable from above, unstable from below.

It reaches zero in finite time. With $u = \tfrac{3}{2} - y$ , the equation becomes $\dfrac{du}{dt} = 2u^2$ , so $\dfrac{1}{u} = \dfrac{1}{u_0} - 2t$ with $u_0 = 0.1$ . The pen is empty when $y = 0$ , i.e. $u = \tfrac{3}{2}$ :

\tfrac{2}{3} = 10 - 2t \quad\Longrightarrow\quad t = \tfrac{14}{3} \approx 4.67

Problem 3 · Moderate Harvesting

A pen follows $\dfrac{dy}{dt} = 6y - 2y^2 - 4$ . Find its critical points and their types, then decide which starting populations end at a stable level.

Critical points

Starting populations that end at a stable level

✅ Correct! The lower critical point

y = 1

is the critical population: start above it and the stock ends at

y = 2

❌ Those are the crossings with no harvest.

0

and

3

solve

6y - 2y^2 = 0

. Subtracting

h = 4

lowers the curve, and both crossings move inward.

❌ Check the sign between the roots. At

y = \tfrac{3}{2}

the rate is

9 - \tfrac{9}{2} - 4 = \tfrac{1}{2} > 0

, so populations between

1

and

2

rise — away from

1

and toward

2

❌ Sign slip in the factoring.

y^2 - 3y + 2 = (y - 1)(y - 2)

: the roots must add to

3

and multiply to

2

❌ Too strict. Populations between

1

and

2

also make it: there

\tfrac{dy}{dt} > 0

, so they rise to

2

❌ Falling is not dying. A population above

2

decreases, but only down to the stable level

2

, where it levels off.

❌ Too generous. Below the critical population

1

the rate is negative and there is no level to settle at, so those stocks reach zero.

❌ Not quite. Solve

2y^2 - 6y + 4 = 0

and read the sign of the rate between the roots.

❌ Not quite. Follow the arrows: which starting heights move toward

y = 2

Show solution

Here $h = 4 < h_m = \tfrac{9}{2}$ , so there are two critical points. Factor the right-hand side:

6y - 2y^2 - 4 = -2\left(y^2 - 3y + 2\right) = -2(y - 1)(y - 2)

The quadratic formula agrees: $y = \dfrac{6 \pm \sqrt{36 - 32}}{4} = \dfrac{6 \pm 2}{4}$ , giving $1$ and $2$ .

The rate is positive between $1$ and $2$ and negative outside, so $y = 1$ is unstable and $y = 2$ is stable.

$y(0) > 2$ : decreases to $2$ .
$1 < y(0) < 2$ : increases to $2$ .
$y(0) = 1$ : stays at the unstable level $1$ .
$y(0) < 1$ : decreases and reaches $0$ .

So exactly the stocks with $y(0) > 1$ end at the stable level $2$ ; $y = 1$ is the critical population.

Problem 4 · Setting a Safe Quota

For $\dfrac{dy}{dt} = ay - by^2 - h$ , a manager wants the two critical points at least 20% of the carrying capacity $\tfrac{a}{b}$ apart, so that a stock on the stable level has room to fall in a bad season. What is the largest harvest that allows this?

✅ Correct! A margin of 20% of the capacity costs only a 4% cut in the harvest, because

\sqrt{0.04} = 0.2

❌ The gap is not proportional to the cut. It grows like a square root: a 20% cut already opens a gap of

\sqrt{0.2} \approx 45\%

of the capacity. A 20% gap needs a cut of only

0.2^2 = 4\%

❌ That is half the gap.

\dfrac{\sqrt{a^2 - 4bh}}{2b}

is the distance from the peak to one critical point; the two critical points are

\dfrac{\sqrt{a^2 - 4bh}}{b}

apart.

❌ The square root sits on the cut, not on $h$ . The gap is

\sqrt{1 - h/h_m}

times the capacity, not

1 - \sqrt{h/h_m}

times it.

❌ Not quite. Write the distance between the two critical points and compare it with

0.2 \cdot \tfrac{a}{b}

Show solution

The critical points $\dfrac{a \pm \sqrt{a^2 - 4bh}}{2b}$ are $\dfrac{\sqrt{a^2 - 4bh}}{b}$ apart. Require that gap to be at least $0.2 \cdot \tfrac{a}{b}$ :

\frac{\sqrt{a^2 - 4bh}}{b} \ge \frac{0.2\,a}{b} \iff a^2 - 4bh \ge 0.04\,a^2 \iff h \le 0.96 \cdot \frac{a^2}{4b}

So the largest allowed harvest is $h = 0.96\,h_m$ . Equivalently, the gap is $\sqrt{1 - h/h_m}$ times the capacity, and $\sqrt{1 - 0.96} = 0.2$ .

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