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Cows on the Meadow

Newton's grazing problem, animated

Drag cows onto a meadow that regrows daily. Each day, +g tiles bloom, then N cows crop the grass — so the net change Δ = g − N becomes a meter you can watch breathe. Four levels build the cow grazing problem (牛吃草问题) from free play to a leaky water tank, the same model wearing a different skin.

What this game shows · Cow Grazing Problem (牛吃草问题)

Newton's grazing problem looks confusing because two things happen at once: grass grows while cows eat. This animated game runs both rates at the same time so you can read the daily net change Δ = g − N straight from the meter — no algebra required.

Daily growth (g): tiles added to the meadow each day.
Cows (N): tiles eaten each day, one per cow.
Net change (Δ): Δ = g − N. Negative kills the field; zero sustains it forever.

Aligned with CCSS 6.EE.C.9 (two related quantities) and 6.RP.A.2 (unit rate). Recommended for Grade 6, with a level 4 transfer story for Grade 7.

Free Range

Drag cows and daily growth. Press ▶ Day +1 and watch the meadow rise or fall — the panel writes Δ = g − N for you.

Hint · Try cows = 0, growth = 5: the meadow fills up. Then cows = 10, growth = 5: it empties.

Meadow · 100 tiles

60/ 100

Day

Grass-o-meterΔ = 5 − 4 = +1

+5 grow/day−4 eaten/day Δ = +1

What the meadow says

Daily growth → +g = +5
Daily eaten → −N = −4
Net per day → Δ = g − N = 5 − 4 = +1
After 0 days → 60 + 0 × (+1) = 60

Cows N · 4 Daily growth g · 5 Initial grass · 60

FAQ

Cow grazing, answered.

Tap to expand

01 What is Newton's cow grazing problem (牛吃草问题)? Olympiad

A classic Olympiad puzzle where grass grows daily at rate g while N cows eat one tile each. Two rates run at once, so the meadow holds out for as long as the original supply plus accumulated growth covers consumption.

02 How do you find the daily growth rate from two scenarios? g = 15

Subtract cow-day totals. If 27 cows × 6 days and 23 cows × 9 days both finish the meadow, then 207 − 162 = 45 extra cow-days came from 3 extra days of growth, so g = 15. Original supply = 162 − 6 × 15 = 72.

03 What is the maximum sustainable cow count? N ≤ g

Exactly N = g. Below that, grass piles up against the cap; above it, the meadow eventually dies. Δ = g − N is the daily net change — set Δ ≥ 0 for sustainability.

04 Which grade is this game for? Grade 6

Grade 6, aligned with CCSS 6.EE.C.9 (two related quantities) and 6.RP.A.2 (unit rate). The leaky-tank transfer level reuses the same model for Grade 7 ratio applications.

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