When does a logistic population grow fastest?
You will be able to: Find the maximum growth rate and determine whether a starting population reaches it in forward time.
When does a logistic population grow fastest?
In a habitat with capacity 100, a population of 10 has plenty of resources but few organisms. A population of 90 has many organisms but scarce resources. The largest total growth rate occurs halfway between.
A useful starting point: How does a carrying capacity change proportional growth? →
Words and symbols before equations
- Growth-rate function G(P)
- The rate expressed as a function of the population.
- Inflection point
- A point where the population curve changes concavity.
- Maximum positive rate
- The largest value of P′, distinct from the largest population.
- Forward-time condition
- Whether the solution reaches a population level at t≥0.
What this picture assumes
Original model; rounded readouts and finite sampled graphs. P′=0.5P(1−P/100), t in days. K=100 and r=0.5/day fixed; P is treated continuously. Teal dashed line is capacity. Exact curves include separate 0 and 100 equilibria. The rate graph distinguishes population from growth rate.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- P₀=20; initial rate=8 organisms/day. At 6 days P=83.3925. Reaches 50 at t=2.77259 days; peak positive rate 12.5 organisms/day.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
For fixed r>0 and K>0, G(P)=rP−(r/K)P² is a downward-opening parabola. Its derivative with respect toP is r−2rP/K, zero at P=K/2.
Thus the maximum positive rate is rK/4. In the alternative notation kP(K−P), it is kK²/4. Do not confuse the population K/2 with this rate.
Along a solution, P″=r(1−2P/K)P′. For 0<P<K, P′>0, so the curve is concave up below K/2 and concave down above K/2.
If 0<P₀<K/2, the solution reachesK/2 later and has an inflection there. If K/2<P₀<K, it is already past that point; its largest forward-time growth rate is its initial rate. AboveK the population decreases and does not passK/2 while approachingK.
A worked example, step by step
With r=0.5 day⁻¹, K=100 and P₀=20, find the peak growth rate and when it occurs using the displayed solution.
- Peak growth occurs at P=K/2=50 organisms.
- The rate there is rK/4=12.5 organisms/day.
- The solution is P=100/[1+4e^(−0.5t)]. SetP=50 to obtain 4e^(−0.5t)=1.
- Thus t=2ln 4≈2.773 days. Because 20<50, this time is positive and belongs to the forward evolution.
The fastest population growth is not at carrying capacity, where the rate is zero. Check the initial population before claiming a future inflection.
IfP₀=80 andK=100, will the forward solution pass 50?
Compare with an explanation
No. It increases from 80 toward 100, so its largest future rate occurs at the start.
Predict. Change one thing. Explain.
Compare initial populations 20,50 and 80. Read the rate-versus-population parabola and the forward-time peak message. Explain why only the first reaches its peak strictly after t=0.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
P₀=20; initial rate=8 organisms/day. At 6 days P=83.3925. Reaches 50 at t=2.77259 days; peak positive rate 12.5 organisms/day.
Original model; rounded readouts and finite sampled graphs. P′=0.5P(1−P/100), t in days. K=100 and r=0.5/day fixed; P is treated continuously. Teal dashed line is capacity. Exact curves include separate 0 and 100 equilibria. The rate graph distinguishes population from growth rate.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the rate units, derivative signs, initial condition, domain or solution check. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionFor P′=0.01P(60−P), find the population and rate at maximum growth. Explain whetherP₀=40 reaches that population in the future.
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Compare with the answer and four-point rubric
- 1 point: The capacity isK=60.
- 1 point: The maximum positive rate occurs at P=30.
- 1 point: Evaluate 0.01(30)(30)=9 organisms per time unit.
- 1 point: Starting at 40, the solution increases toward 60, so it never reaches 30 at t≥0; its maximum forward growth rate is its initial 8.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Where is logistic positive growth fastest?
At half capacity, P=K/2.
RECALL 2What is the growth rate at carrying capacity?
Zero.
RECALL 3Why must initial conditions enter peak reasoning?
The trajectory may already have passed half capacity or may never reach it.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
When does a logistic population grow fastest?
- Peak positive rate at P=K/2: P′=rK/4.
- P″=r(1−2P/K)P′.
Remember: The fastest population growth is not at carrying capacity, where the rate is zero. Check the initial population before claiming a future inflection.
Conditions: Original model; rounded readouts and finite sampled graphs. P′=0.5P(1−P/100), t in days. K=100 and r=0.5/day fixed; P is treated continuously. Teal dashed line is capacity. Exact curves include separate 0 and 100 equilibria. The rate graph distinguishes population from growth rate.
Refresh Kid · AP Calculus BC Unit 7 · Objectives FUN-7.H · Review edition
Framework, scope and review status
Mapped to College Board CED Topic 7.9, FUN-7.H. CED and Fall 2026 clarifications checked September 17, 2026. All nine BC topics are included, with Euler’s method and logistic interpretation. The correction to FUN-7.B.2 is reflected: an equation may have infinitely many solutions. Focused explanations, examples, visual models and practice are original Refresh Kid work.
Rate and amount units, signs, initial data, lost equilibria, solution intervals and model assumptions are explicit. Slope fields and Euler polygons are finite illustrations; exact algebra supports identities and limits. The logistic explicit formula supports the visualization; required interpretation can be done from the rate law without deriving that formula. The smooth examples have unique solutions; no universal uniqueness claim is made.
Organic Chemistry Tutor video titles and destinations were checked, not the full videos. Khan Academy’s destination was checked; its JavaScript lesson contents were not fully readable by the research tool. OpenStax sections 4.1–4.4 supplied conceptual cross-checks. No provider questions, artwork or scripts were copied. No affiliation or endorsement is implied.
GitHub’s 3D website collection informed optional spatial inspection and camera controls. The original tank reuses self-hosted Three.js with its MIT license retained. The 2×2 dm base connects water depth to volume: 1 dm³=1 L. This tank uses a feedback-controlled pump, not a gravity-drain law. Its 3D height and labeled 2D graphs use the same exact exponential formula. Rotation changes only the view, and complete explanations remain available without WebGL.
Independent teacher review and student usability testing remain pending. Technical checks do not certify mathematical accuracy, full accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus BC questions and scoring guides. This multi-unit archive is not assigned as a complete Unit 7 task.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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