Optimisation: 7 exam questions solved step by step

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What this lesson covers

  • An optimisation problem is solved by writing the quantity as a function of one variable and finding its turning point.
  • The answer must be shown to be a maximum or minimum, and must make sense in context.
Key idea

7 questions, one method

  • 1. Write the quantity in one variable.
  • 2. Differentiate and set it to 0.
  • 3. Show it is a max or a min.
  • 4. Answer the question, with units.
Your turn

Question 1 · Achieved, in context

  • A school fundraiser sells tickets at $x.
  • Profit: P = −5x² + 80x − 120 dollars
  • Find the price for the greatest profit,
  • and the greatest profit.
Worked example

Q1: the best ticket price

  • Set P′ = 0.
  • The shape proves a max.
  1. P′ = −10x + 80differentiate
  2. −10x + 80 = 0, so x = 8stationary
  3. −5 < 0: opens downwardso a maximum
  4. P = −5(8)² + 80(8) − 120x = 8
  5. = −320 + 640 − 120 = 200
  6. $8 a ticket, profit $200
Your turn

Question 2 · Achieved, in context

  • Packhouse cost per tray, x workers:
  • C = 2x² − 48x + 400 cents
  • How many workers give the least cost?
  • What is that cost, in dollars?
Worked example

Q2: the cheapest tray

  • Gradient either side.
  • Cents to dollars.
  1. C′ = 4x − 48differentiate
  2. 4x − 48 = 0, so x = 12stationary
  3. C′(11) = 44 − 48 = −4 < 0falling
  4. C′(13) = 52 − 48 = 4 > 0rising
  5. − then +: a minimum
  6. C = 2(12)² − 48(12) + 400
  7. = 288 − 576 + 400 = 112
  8. 112 cents = $1.12answer in dollars
  9. 12 workers, $1.12 a tray
Your turn

Question 3 · Merit, in context

  • A farmer has 120 m of fencing for a
  • rectangular paddock, split into two pens
  • by one more fence parallel to two sides.
  • Find the greatest total area.
Question 3 · Merit, in context
Worked example

Q3: form the area

  • Name both lengths.
  • Then one variable only.
  1. 3x + 2y = 120three x, two y
  2. y = 60 − 1.5x− 3x, then ÷ 2
  3. A = xy = x(60 − 1.5x)length × width
  4. A = 60x − 1.5x²expand
Worked example

Q3: the greatest area

  • Check either side.
  • Answer with units.
  1. A = 60x − 1.5x²
  2. A′ = 60 − 3xdifferentiate
  3. 60 − 3x = 0, so x = 20stationary
  4. A′(19) = 3 > 0, A′(21) = −3 < 0
  5. + then −: a maximum
  6. y = 60 − 1.5 × 20 = 30
  7. A = 20 × 30 = 600 m²
  8. 20 m by 30 m, area 600 m²
Your turn

Question 4 · Merit, in context

  • A Rotorua tour company runs x tours a week.
  • Profit: P = −x³ + 6x² + 15x
  • in hundreds of dollars.
  • Find the greatest weekly profit.
Worked example

Q4: the tour company

  • Reject what cannot be.
  • P is in hundreds.
  1. P′ = −3x² + 12x + 15differentiate
  2. = −3(x² − 4x − 5)take out −3
  3. = −3(x − 5)(x + 1) = 0factorise
  4. x = 5 or x = −1
  5. x = −1tours cannot be negative
  6. P′(4) = 15 > 0, P′(6) = −21 < 0
  7. P = −125 + 150 + 75 = 100x = 5
  8. 100 × $100 = $10,000hundreds of dollars
  9. 5 tours a week: $10,000 profit
Your turn

Question 5 · Merit, in context

  • A box is made from a 16 cm by 10 cm card
  • by cutting a square of side x cm from
  • each corner and folding up the sides.
  • Find the greatest volume.
Question 5 · Merit, in context
Worked example

Q5: form the volume

  • Base loses 2x each way.
  • Expand fully.
  1. length 16 − 2x, width 10 − 2x
  2. height x
  3. V = x(16 − 2x)(10 − 2x)length × width × height
  4. (16 − 2x)(10 − 2x)
  5. = 160 − 32x − 20x + 4x²
  6. = 160 − 52x + 4x²collect
  7. V = 4x³ − 52x² + 160xmultiply by x
Worked example

Q5: the greatest volume

  • Reject the root that fails.
  • Answer with units.
  1. V = 4x³ − 52x² + 160x
  2. V′ = 12x² − 104x + 160differentiate
  3. = 4(3x² − 26x + 40)take out 4
  4. = 4(3x − 20)(x − 2) = 0factorise
  5. x = 2 or x = 20/3 ≈ 6.67
  6. x = 20/3: 10 − 40/3 < 0negative width
  7. V′(1) = 68 > 0, V′(3) = −44 < 0
  8. V = 2 × 12 × 6 = 144 cm³x = 2
  9. cut 2 cm squares: 144 cm³
Your turn

Question 6 · Excellence

  • A wire frame for a box with a square base
  • uses 60 cm of wire for its 12 edges.
  • Find the largest volume the box can have.
Worked example

Q6: form the volume

  • Two unknowns: link them.
  • Then one variable.
  1. base x by x, height h
  2. 8x + 4h = 608 base edges, 4 upright
  3. 4h = 60 − 8x
  4. h = 15 − 2x÷ 4
  5. V = x² × h = x²(15 − 2x)base area × height
  6. V = 15x² − 2x³expand
Worked example

Q6: the largest volume

  • Reject x = 0.
  • Prove the maximum.
  1. V = 15x² − 2x³
  2. V′ = 30x − 6x²differentiate
  3. 6x(5 − x) = 0factorise
  4. x = 0 or x = 5
  5. x = 0no box at all
  6. V′(4) = 120 − 96 = 24 > 0rising
  7. V′(6) = 180 − 216 = −36 < 0falling
  8. h = 15 − 10 = 5, V = 5² × 5
  9. a 5 cm cube, 125 cm³
Your turn

Question 7 · Excellence, in context

  • A Wellington ferry charges $40 a ticket
  • and sells 600 tickets a day. Each $1 rise
  • in price loses 10 sales a day.
  • What price gives the most revenue?
Worked example

Q7: form the revenue

  • x = number of $1 rises.
  • Revenue = price × sales.
  1. price = 40 + x dollars
  2. tickets = 600 − 10x
  3. R = (40 + x)(600 − 10x)price × tickets
  4. = 24,000 − 400x + 600x − 10x²
  5. R = 24,000 + 200x − 10x²collect
Worked example

Q7: the best price

  • x is not the price.
  • Answer the question asked.
  1. R = 24,000 + 200x − 10x²
  2. R′ = 200 − 20xdifferentiate
  3. 200 − 20x = 0, so x = 10stationary
  4. R′(9) = 20 > 0, R′(11) = −20 < 0
  5. + then −: a maximum
  6. price = 40 + 10 = $50x is not the price
  7. tickets = 600 − 100 = 500
  8. R = 50 × 500 = $25,000
  9. $50 a ticket, $25,000 a day
Summary

The method that solved all seven

  • Form, solve, reject, prove, then answer the question.
StepHow
Namethe variable, with units
Linkuse the constraint
Formone quantity, one variable
Solvederivative = 0
Rejectroots that cannot happen
Provegradient either side, or shape
Answerthe question asked, with units

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