Lesson 8.1 — Build a Growing-Resource Model
牛吃草问题：建立“初始量—增长—消耗”模型
A pasture already has grass. More grass grows every day. Cows eat every day. Learn to keep all three parts in one clear mathematical model.
Meet the three moving parts
A growing-resource problem is different from an ordinary “use up a fixed pile” problem. The amount changes while it is being used.
The pasture story
At the beginning, a pasture already contains grass. Every day, new grass grows. At the same time, cows eat grass.
already present
added each day
removed each day
Identify each part
This demonstration uses 100 starting units, 10 units of growth per day, and 12 units of consumption per day.
Choose one useful unit: the cow-day
We will measure amounts of grass by how much one cow eats in one day.
Definition
One cow-day is the amount of grass eaten by one cow in one day.
total consumption = number of cows × number of daysFor example, 6 cows eating for 4 days use:
6 × 4 = 24 cow-days of grassEvery cell is one cow eating for one day.
Count the consumption units
Build the first pasture scenario
A pasture feeds 15 cows for 20 days. At the end, all available grass has been eaten.
What the cows used
15 × 20 = 300 cow-daysThose 300 units came from two places:
- grass present at the beginning;
- grass that grew during the 20 days.
Using letters:
S + 20g = 300S means starting grass. g means daily growth.
The two colored parts together equal everything the cows consumed.
Complete the scenario
Build the second pasture scenario
In a separate trial, the pasture starts again with the same amount of grass and has the same daily growth rate. This time, 20 cows eat all available grass in exactly 10 days.
Same pasture, shorter time
20 × 10 = 200 cow-daysThe 200 units again consist of:
starting grass + 10 days of growth = 200S + 10g = 200This row contains less new growth because only 10 days pass.
Complete the second scenario
Subtract the scenarios and cancel the shared start
The unknown starting amount appears in both equations. Subtraction removes it.
- Subtract the second equation from the first.
- The identical S terms cancel.
- Only the extra 10 days of growth remain.
Only the difference in growth and the difference in consumption remain.
Finish the subtraction
Recover the starting grass
Now that the daily growth is known, remove all growth from either scenario.
Use the 20-day scenario
S = 300 − 20 × 10S = 300 − 200 = 100Check with the 10-day scenario
S = 200 − 10 × 10S = 200 − 100 = 100Complete both checks
Find the net rate that reduces the starting stock
Daily growth replaces some of what the cows eat. Only the excess consumption reduces the original grass.
The net-depletion rule
net depletion per day = cow consumption − grass growthFor 15 cows:
15 − 10 = 5 units per day100 ÷ 5 = 20 daysFor 20 cows:
20 − 10 = 10 units per day100 ÷ 10 = 10 days- daily use > daily growth: the original grass eventually runs out;
- daily use = daily growth: the original amount stays unchanged;
- daily use < daily growth: the total grass increases.
Fifteen cows use 5 more units per day than the pasture grows.
Use the net rate
Build and test the complete model
Before the pasture runs out, remaining grass = S + gt − nt. At the exact time it runs out, the remaining amount is zero.
The equation at the emptying time
S + gt = ntwhere:
- S = starting stock;
- g = growth per day;
- n = consumption per day (one cow uses one cow-day unit per day);
- t = number of days.
For a positive starting stock S and daily use n greater than growth g, rearranging gives the emptying time:
S = (n − g)tt = S ÷ (n − g)Check the model
Resource-model laboratory
A fresh pasture — independent investigation
Try the eight steps yourself, using a small hint if needed. Check an attempt before opening worked review. Correct all eight to complete this investigation.
In two separate trials, a pasture runs out of grass in exactly 12 days with 18 cows or exactly 20 days with 14 cows. Both trials begin with the same amount of grass. Growth is constant and all cows eat equally. Use cow-day units.
Write quantities in one consistent unit. Enter decimals or fractions such as 30/7; a decimal within 0.001 is accepted.
Hint
Multiply rate by time.
Worked solution
Multiply 18 × 12 = 216 cow-days.
Hint
Compare totals only after using the same unit.
Worked solution
Multiply 14 × 20 = 280 cow-days.
Hint
Subtract the shorter trial’s 12 days from the longer trial’s 20 days.
Worked solution
20 − 12 = 8 days.
Hint
Subtract the shorter trial’s total consumption from the longer trial’s total consumption. Their identical starting amounts cancel.
Worked solution
280 − 216 = 64. The identical starting grass cancels.
Hint
Divide the extra grass by the extra days of growth.
Worked solution
64 ÷ 8 = 8 cow-day units per day.
Hint
Subtract 12 days of growth from the 216 units consumed in the first trial.
Worked solution
216 − 12 × 8 = 120 cow-days. Check: 280 − 20 × 8 = 120.
Hint
Subtract daily growth from the 20 cows’ daily consumption.
Worked solution
20 − 8 = 12 cow-day units per day.
Hint
Divide remaining stock by its net depletion rate; check the time or capacity requested.
Worked solution
120 ÷ 12 = 10 days. Check: 120 + 8 × 10 = 20 × 10.
Exit ticket
Answer all five questions correctly. Complete the other core missions and the independent investigation to earn the certificate.
Lesson checkpoints completed
Growing-Resource Model Builder
You completed the checkpoints on starting stock, growth and consumption. Revisit any steps for which you needed solution help. The model is S + gt = nt, with net-depletion rate n − g.
Lesson 8.1 completed
Teaching notes
The interactive pasture tank, cow-day array, cancellation animation, net-rate explorer, model laboratory, feedback, and assessments are added teaching scaffolds.