riaology
Unit 3 · Lesson 6 of 7

A constraint turns a design question into one variable.

A river supplies one side of an enclosure. Decide how to distribute a fixed amount of fencing to maximize useful area.

Change the input. Follow the mathematics.

What you'll be able to do
  • Set up the model and calculate its area at the chosen dimension.
  • Connect the equation to the changing visual.
  • Explain the assumptions and identify a misleading interpretation.
Predict before exploring

What will change if you increase fence allocated to each short side? Predict the direction, then explain the response.

A model you can changeMove one control at a time.
Rectangle sides 7.5 and -2.5 meters; geometry changes at constant perimeter7.5 m15 mBoth dimensions respond to x
Short side x (m) versus Enclosed area (m²). Area. Selected area at the chosen dimension: 112.5 m². The maximum is 112.5 m² at x=7.5 m. Derive the constraint before differentiating the objective.0056.257.5112.515Enclosed area (m²)Short side x (m)
Area

area at the chosen dimensionA(7.5)=7.5(30−2(7.5))=112.5 m2,x∗=7.5 mA(7.5)=7.5(30-2(7.5))=112.5\,\mathrm{m^2},\quad x_*=7.5\,\mathrm m

The maximum is 112.5 m² at x=7.5x=7.5 m. Derive the constraint before differentiating the objective.

Assumptions, units, and limits

A river forms one side, so 30 m of fence encloses only three sides. Feasible x is between 0 and 15 m.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

01

From the situation to an equation

Write the perimeter constraint first and eliminate one dimension. Differentiate the resulting area function, test the feasible endpoints, and verify the critical point is a maximum. The physical domain excludes negative side lengths.

2x+y=30,A=x(30−2x),A′=30−4x=02x+y=30,\quad A=x(30-2x),\quad A'=30-4x=0
02

Keep the reasoning visible

A′′=−4<0A''=-4<0

The maximum is 112.5 m² at x=7.5x=7.5 m. Derive the constraint before differentiating the objective.

03

What makes this model useful?

A river forms one side, so 30 m of fence encloses only three sides. Feasible x is between 0 and 15 m.

A different fencing arrangement creates a different constraint. Four-sided fencing would have a different optimum.

A worked example

Calculate the starting scenario

  1. Use fence allocated to each short side = 7.5 m.
  2. A(7.5)=7.5(30−2(7.5))=112.5 m2,x∗=7.5 mA(7.5)=7.5(30-2(7.5))=112.5\,\mathrm{m^2},\quad x_*=7.5\,\mathrm m
  3. The maximum is 112.5 m² at x=7.5x=7.5 m. Derive the constraint before differentiating the objective. The checked quantity is area at the chosen dimension.
Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the area at the chosen dimension using fence allocated to each short side 7.5 m. State the meaning of its sign and units.

Checked quantity: area at the chosen dimension · m². Follow the definition for finite-cutoff and numerical-model questions.

Hint

Write the perimeter constraint first and eliminate one dimension. Differentiate the resulting area function, test the feasible endpoints, and verify the critical point is a maximum. The physical domain excludes negative side lengths.

Explore the visual solution
Rectangle sides 7.5 and -2.5 meters; geometry changes at constant perimeter7.5 m15 mBoth dimensions respond to x
Short side x (m) versus Enclosed area (m²). Area. Selected area at the chosen dimension: 112.5 m². The maximum is 112.5 m² at x=7.5 m. Derive the constraint before differentiating the objective.0056.257.5112.515Enclosed area (m²)Short side x (m)
Area

area at the chosen dimensionA(7.5)=7.5(30−2(7.5))=112.5 m2,x∗=7.5 mA(7.5)=7.5(30-2(7.5))=112.5\,\mathrm{m^2},\quad x_*=7.5\,\mathrm m

The maximum is 112.5 m² at x=7.5x=7.5 m. Derive the constraint before differentiating the objective.

Assumptions, units, and limits

A river forms one side, so 30 m of fence encloses only three sides. Feasible x is between 0 and 15 m.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Write the perimeter constraint first and eliminate one dimension. Differentiate the resulting area function, test the feasible endpoints, and verify the critical point is a maximum. The physical domain excludes negative side lengths.

2x+y=30,A=x(30−2x),A′=30−4x=02x+y=30,\quad A=x(30-2x),\quad A'=30-4x=0
Calculate and explain

Apply the idea to a new scenario

Allocate x=5x=5 m to each short side; find enclosed area and compare with the optimum.

Checked quantity: area at the chosen dimension · m². Follow the definition for finite-cutoff and numerical-model questions.

Hint

Write the perimeter constraint first and eliminate one dimension. Differentiate the resulting area function, test the feasible endpoints, and verify the critical point is a maximum. The physical domain excludes negative side lengths.

Explore the visual solution
Rectangle sides 5 and 0 meters; geometry changes at constant perimeter5 m20 mBoth dimensions respond to x
Short side x (m) versus Enclosed area (m²). Area. Selected area at the chosen dimension: 100 m². The maximum is 112.5 m² at x=7.5 m. Derive the constraint before differentiating the objective.0056.257.5112.515Enclosed area (m²)Short side x (m)
Area

area at the chosen dimensionA(5)=5(30−2(5))=100 m2,x∗=7.5 mA(5)=5(30-2(5))=100\,\mathrm{m^2},\quad x_*=7.5\,\mathrm m

The maximum is 112.5 m² at x=7.5x=7.5 m. Derive the constraint before differentiating the objective.

Assumptions, units, and limits

A river forms one side, so 30 m of fence encloses only three sides. Feasible x is between 0 and 15 m.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Write the perimeter constraint first and eliminate one dimension. Differentiate the resulting area function, test the feasible endpoints, and verify the critical point is a maximum. The physical domain excludes negative side lengths.

2x+y=30,A=x(30−2x),A′=30−4x=02x+y=30,\quad A=x(30-2x),\quad A'=30-4x=0
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Rectangle sides 7.5 and -2.5 meters; geometry changes at constant perimeter7.5 m15 mBoth dimensions respond to x
Short side x (m) versus Enclosed area (m²). Area. Selected area at the chosen dimension: 112.5 m². The maximum is 112.5 m² at x=7.5 m. Derive the constraint before differentiating the objective.0056.257.5112.515Enclosed area (m²)Short side x (m)
Area

area at the chosen dimensionA(7.5)=7.5(30−2(7.5))=112.5 m2,x∗=7.5 mA(7.5)=7.5(30-2(7.5))=112.5\,\mathrm{m^2},\quad x_*=7.5\,\mathrm m

The maximum is 112.5 m² at x=7.5x=7.5 m. Derive the constraint before differentiating the objective.

Assumptions, units, and limits

A river forms one side, so 30 m of fence encloses only three sides. Feasible x is between 0 and 15 m.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Write the perimeter constraint first and eliminate one dimension. Differentiate the resulting area function, test the feasible endpoints, and verify the critical point is a maximum. The physical domain excludes negative side lengths.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

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