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Measurement and Geometry

Farm data: palay yield, inputs and break-even

Heometriya at pagsukat

Datos sa bukid: ani ng palay, gastos at break-even

Hectares, tonnes and pesos: convert units, compare provincial yields and find the farmgate price at which a rice farmer stops losing money.

Where it fits

Competencies

  • gm11.t1.w6.c2 Convert measurements with different units in the same system and between different systems.
  • gm11.t1.w10.c2 Apply the functions to model and solve real-life problems.
  • gm11.t2.w2.c3 Identify the appropriate measure of central tendency and measures of variability of a given data set for different purposes.

The data

Palay production, selected provinces (one season)
ProvinceYield (t/ha)Farmgate price (₱/kg)Cost per hectare (₱)Profit per hectare (₱)
Nueva Ecija4.9₱19.50₱64,000₱31,550
Isabela4.5₱18.00₱60,000₱21,000
Pangasinan4.2₱19.00₱62,000₱17,800
Iloilo3.8₱20.00₱58,000₱18,000
Camarines Sur3.6₱18.50₱55,000₱11,600
Yield (t/ha) by Province0 t/ha2.5 t/ha4.9 t/haNueva Ecija: 4.9Isabela: 4.5Pangasinan: 4.2Iloilo: 3.8Camarines Sur: 3.6Nueva EcijaIsabelaPangasinanIloiloCamarines S…
Yield (t/ha) by Province

Source: Representative 2026 figures for classroom use; check current official rates before using for real decisions. Yields and prices are representative of PSA and DA reports for irrigated palay.

Scenario and worked example

Tatay Lando's season

Engage

Tatay Lando farms 1.5 hectares in Pangasinan. His cooperative quotes ₱62,000 per hectare for seed, fertiliser, labour and irrigation. A trader offers ₱15 per kilo at harvest; the NFA buys at ₱23 but with delays. He expects 4.2 t/ha. Should he sell to the trader?

Worked example: revenue and profit

Explain
Yield=4.2×1.5=6.3 t=6,300 kg\text{Yield} = 4.2 \times 1.5 = 6.3\text{ t} = 6{,}300\text{ kg}Cost=62,000×1.5=93,000\text{Cost} = 62{,}000 \times 1.5 = 93{,}000Trader: 6,300×15=94,500  ⇒  profit 1,500\text{Trader: } 6{,}300 \times 15 = 94{,}500 \;\Rightarrow\; \text{profit } 1{,}500NFA: 6,300×23=144,900  ⇒  profit 51,900\text{NFA: } 6{,}300 \times 23 = 144{,}900 \;\Rightarrow\; \text{profit } 51{,}900

Break-even as a function

Elaborate

Revenue per hectare is $R(p) = 4000p$ and cost is the constant $C = 62{,}000$. The graphs cross at the break-even price.

4000p=62,000⇒p=15.504000p = 62{,}000 \Rightarrow p = 15.50

Units matter

Explain
  • 1 hectare = 10,000 m² (about 1.4 football pitches).
  • 1 tonne = 1,000 kg; a sack of palay is about 50 kg.
  • Yield × area × price = revenue; keep units consistent.

Discuss in class

  1. Which province earns the most profit per hectare? Is it the one with the highest yield?
  2. How much does a ₱1 change in farmgate price change Tatay Lando's profit?
  3. What risks (weather, pests, delays) does the arithmetic ignore?

Problems

5 problems · check your answer, then open the solution

  1. Problem 1 Basic

    A farm lot measures 2.5 hectares. Express the area in square metres.

    Show step-by-step solution
    1. 1 ha=10,000 m21\text{ ha} = 10{,}000\text{ m}^2.
    2. 2.5×10,000=25,000 m22.5 \times 10{,}000 = 25{,}000\text{ m}^2.

    Answer: 25000 m²

  2. Problem 2 Standard

    A farmer harvests 4.2 tonnes of palay per hectare on 1.5 hectares and sells at ₱20 per kilogram. Find the gross revenue.

    Show step-by-step solution
    1. Total yield: 4.2×1.5=6.34.2 \times 1.5 = 6.3 tonnes =6,300= 6{,}300 kg.
    2. Revenue: 6,300×20=126,0006{,}300 \times 20 = 126{,}000 pesos.

    Answer: 126000 ₱

  3. Problem 3 Standard

    Producing one hectare of palay costs ₱62,000 and yields 4,000 kg. What farmgate price per kilogram lets the farmer break even?

    Show step-by-step solution
    1. Break-even when revenue equals cost: 4,000p=62,0004{,}000p = 62{,}000.
    2. p=62,000÷4,000=15.50p = 62{,}000 \div 4{,}000 = 15.50 pesos per kilogram.
    3. Any price above ₱15.50 gives profit; the revenue function R(p)=4000pR(p)=4000p is linear in price.

    Answer: 15.5 ₱

  4. Problem 4 Standard

    Average palay yields (tonnes per hectare) in five provinces: 4.2, 3.8, 4.5, 3.6, 4.9. Find the mean yield.

    Show step-by-step solution
    1. Sum: 4.2+3.8+4.5+3.6+4.9=21.04.2+3.8+4.5+3.6+4.9 = 21.0.
    2. Mean =21.0÷5=4.2= 21.0 \div 5 = 4.2 tonnes per hectare.

    Answer: 4.2 t/ha

  5. Problem 5 Challenge Non-routine

    Urea fertiliser rose from ₱1,400 to ₱1,750 per bag. A farmer uses 8 bags per hectare on 3 hectares. By how many pesos did the fertiliser cost increase, and what is the percentage increase? Enter the peso increase.

    Show step-by-step solution
    1. Increase per bag: 1,750−1,400=3501{,}750 - 1{,}400 = 350; percentage 350/1,400=25%350/1{,}400 = 25\%.
    2. Bags used: 8×3=248 \times 3 = 24.
    3. Extra cost: 24×350=8,40024 \times 350 = 8{,}400 pesos.

    Answer: 8400 ₱

Graph revenue and cost lines More practice: Units and ConversionMore practice: Functions and Their GraphsMore practice: Data and Descriptive Statistics

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