NCEA Level 3OptimisationAS 91578

Optimisation: building the model and finding the max or min

What this lesson covers

  • Optimising means writing the quantity as a function of one variable, then finding its stationary point.
  • At Achieved and Merit the question tells you whether it is a maximum or minimum.
Key idea

Optimising: the greatest or least value

  • Differentiate, set to zero, solve.
  • The hard part comes first.
  • Write it as a function of one variable.
Key idea

Four steps

  • 1. Name the variables, and sketch.
  • 2. Write the quantity to optimise.
  • 3. Use the constraint to remove one.
  • 4. Differentiate and set it to zero.
Worked example

40 m of fence, three sides

  • The wall is the fourth side.
  • One variable: the width x.
  1. width = x
  2. length = 40 − 2x2x + length = 40
  3. A = x(40 − 2x)width × length
  4. = 40x − 2x²expand
Worked example

Differentiate and solve

  • Set the derivative to zero.
  • Check: a maximum.
  1. dA/dx = 40 − 4xdifferentiate
  2. 40 − 4x = 0stationary
  3. 4x = 40, so x = 10
  4. d²A/dx² = −4 < 0a maximum
  5. length = 40 − 20 = 20
  6. A = 10 × 20 = 200 m²
Worked example

The graph of the area

  • A parabola, opening down.
  • Zero gradient at x = 10.
  • The top: A = 200.
The graph of the area
Worked example

The 2025 weakness: the model

  • A = xy has two variables.
  • The constraint gives y = 40 − 2x.
  • Substitute: A in terms of x alone.
A = xytwo variablesstuck
2x + y = 40y = 40 − 2xconstraint
A = x(40 − 2x)one variableready
Worked example

An open box from 12 cm card

  • Cut x from each corner.
  • Expand, then differentiate.
  1. height x, base 12 − 2xx off each end
  2. V = x(12 − 2x)²height × base²
  3. = x(144 − 48x + 4x²)expand
  4. = 4x³ − 48x² + 144x
  5. dV/dx = 12x² − 96x + 144differentiate
  6. = 12(x² − 8x + 12)
  7. = 12(x − 2)(x − 6)factorise
Worked example

Solve, reject, check

  • x = 6 makes no box.
  • Second derivative: a max.
  1. dV/dx = 12(x − 2)(x − 6)
  2. 12(x − 2)(x − 6) = 0stationary
  3. x = 2 or x = 6
  4. x = 6: base 12 − 12 = 0no box
  5. d²V/dx² = 24x − 96
  6. x = 2: 48 − 96 = −48 < 0a maximum
  7. V = 2(12 − 4)²
  8. = 2 × 64 = 128 cm³
Worked example

A minimum: an open tank

  • Open top: base, four sides.
  • The volume gives h.
  1. x²h = 32 000volume
  2. h = 32 000/x²
  3. S = x² + 4xhbase + four sides
  4. = x² + 4x(32 000/x²)substitute h
  5. = x² + 128 000/x
  6. = x² + 128 000x⁻¹power form
  7. dS/dx = 2x − 128 000x⁻²differentiate
  8. = 2x − 128 000/x²
Worked example

Solve, check, answer

  • Clear the fraction first.
  • Positive: a minimum.
  1. dS/dx = 2x − 128 000/x²
  2. 2x − 128 000/x² = 0stationary
  3. 2x³ = 128 000multiply by x²
  4. x³ = 64 000, so x = 40
  5. h = 32 000/1600 = 20
  6. d²S/dx² = 2 + 256 000/x³
  7. x = 40: 2 + 4 = 6 > 0a minimum
  8. S = 1600 + 128 000/40
  9. = 1600 + 3200 = 4800 cm²
Key idea

Maximum or minimum?

  • Achieved, Merit: you may assume it.
  • Excellence: be ready to show it.
  • S'' = 6 > 0 at x = 40: a minimum.
Worked example

A rectangle under y = 12 − x²

  • Corners at ±x on the axis.
  • Width 2x, height y.
  1. width 2x, height 12 − x²
  2. A = 2x(12 − x²)
  3. = 24x − 2x³expand
  4. dA/dx = 24 − 6x²
  5. 24 − 6x² = 0stationary
  6. 6x² = 24, so x² = 4
  7. x = 2x > 0, a length
  8. d²A/dx² = −12x = −24 < 0a maximum
  9. A = 4 × 8 = 32width 4, height 8
Key idea

Answer the question asked

  • Greatest area? Give the area, not x.
  • Dimensions? Give both.
  • Say why you rejected a value.
Key idea

Show the model and the derivative

  • No credit for a calculator answer alone.
  • Unknown constants stop the calculator.
  • Use it to check, not to replace.
Key idea

Where the marks are

  • Achieved: differentiate a given model.
  • Merit: build the one-variable model.
  • Excellence: your own model, justified.
Summary

What to take away

  • Build a function of one variable first. The calculus is the short part.
Namethe variables
Writethe quantity
Constraintdown to one variable
Differentiateset it equal to 0
Checkreject impossible x
Answerwhat was asked, units
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • You may assume each is a max or min.
  • Write the model and the derivative.
GivenFind
1rectangle, perimeter 60maximum area
2P = −2x² + 48x − 100maximum profit
318 cm square cardmaximum box volume
4square base, 4000 cm³min surface, open top
Show answer

1. A = x(30 − x), dA/dx = 30 − 2x, x = 15: 225 m², a square. 2. dP/dx = −4x + 48, x = 12: P = 188. 3. V = x(18 − 2x)², dV/dx = 12(x − 3)(x − 9), x = 3 (x = 9 gives no box): 432 cm³. 4. S = x² + 16 000/x, dS/dx = 2x − 16 000/x², x = 20, h = 10: 1200 cm².

Worked example

The answers: questions 1 and 2

  1. 1. 2x + 2y = 60, y = 30 − xperimeter
  2. A = x(30 − x) = 30x − x²
  3. dA/dx = 30 − 2x = 0, x = 15
  4. y = 15: A = 15 × 15 = 225 m²a square
  5. 2. dP/dx = −4x + 48differentiate
  6. −4x + 48 = 0, x = 12
  7. d²P/dx² = −4 < 0a maximum
  8. P = −288 + 576 − 100 = 188
Worked example

The answers: question 3

  1. 3. V = x(18 − 2x)²base 18 − 2x
  2. = 4x³ − 72x² + 324xexpand
  3. dV/dx = 12x² − 144x + 324differentiate
  4. = 12(x − 3)(x − 9) = 0factorise
  5. x = 9: base 18 − 18 = 0no box
  6. x = 3: V = 3(18 − 6)²
  7. = 3 × 144 = 432 cm³
Worked example

The answers: question 4

  1. 4. x²h = 4000, h = 4000/x²volume
  2. S = x² + 4xh = x² + 16 000/xsubstitute h
  3. dS/dx = 2x − 16 000/x² = 0
  4. 2x³ = 16 000, x³ = 8000multiply by x²
  5. x = 20, h = 4000/400 = 10
  6. S = 400 + 16 000/20
  7. = 400 + 800 = 1200 cm²

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