Optimisation: 7 exam questions solved step by step

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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

7 questions, one method

  • 1. Model the quantity from the words.
  • 2. Use the constraint: one variable.
  • 3. Derivative = 0, and solve.
  • 4. Answer the question asked.
Your turn

Question 1 · Achieved, in context

  • A harbour ferry trip costs
  • C = 2v + 800/v dollars at v km/h.
  • Find the speed for the least cost,
  • and that cost. Assume it is a minimum.
Worked example

Q1: the cheapest speed

  • Power form first.
  • A speed is positive.
  1. C = 2v + 800v⁻¹power form
  2. dC/dv = 2 − 800v⁻²differentiate
  3. = 2 − 800/v²
  4. 2 − 800/v² = 0stationary
  5. 2v² = 800, so v² = 400× v²
  6. v = −20a speed is positive
  7. v = 20: C = 40 + 800/20
  8. = 40 + 40 = 80
  9. 20 km/h, costing $80 a trip
Your turn

Question 2 · Achieved, in context

  • A medicine in the blood t hours after
  • a dose: C = 20te⁻ᵗ mg/L.
  • When is C greatest, and what is it?
  • Assume it is a maximum.
Worked example

Q2: the highest concentration

  • Product rule.
  • e⁻ᵗ is never 0.
  1. u = 20t, v = e⁻ᵗproduct rule
  2. u′ = 20, v′ = −e⁻ᵗchain rule
  3. dC/dt = 20e⁻ᵗ − 20te⁻ᵗu′v + uv′
  4. = 20e⁻ᵗ(1 − t)factorise
  5. e⁻ᵗ > 0, so 1 − t = 0stationary
  6. t = 1
  7. C = 20(1)e⁻¹ = 20/e
  8. ≈ 7.36 mg/L3 s.f.
  9. greatest after 1 h: 7.36 mg/L
Your turn

Question 3 · Merit, in context

  • A rectangular paddock of 1800 m² is
  • fenced on three sides; a river is the
  • fourth. Find the least fence needed.
Worked example

Q3: the paddock by the river

  • Area fixes y.
  • Fence: two x, one y.
  1. x out from the river, y along it
  2. xy = 1800, so y = 1800/xarea
  3. L = 2x + y = 2x + 1800/x
  4. dL/dx = 2 − 1800/x²differentiate
  5. 2 − 1800/x² = 0stationary
  6. x² = 900, so x = 30x > 0
  7. y = 1800/30 = 60
  8. L = 60 + 60 = 120
  9. 30 m by 60 m: 120 m of fence
Your turn

Question 4 · Merit, in context

  • A closed food tin is a cylinder
  • holding 128π cm³.
  • Find the radius and height that use
  • the least metal. Assume it is a minimum.
Worked example

Q4: surface area in r only

  • Metal = surface area.
  • Volume gives h.
  1. πr²h = 128πvolume
  2. h = 128/r²÷ πr²
  3. S = 2πr² + 2πrhtwo ends + side
  4. = 2πr² + 2πr(128/r²)substitute h
  5. = 2πr² + 256π/rone variable
Worked example

Q4: the least metal

  • Derivative = 0.
  • Give both dimensions.
  1. S = 2πr² + 256π/r
  2. dS/dr = 4πr − 256π/r²differentiate
  3. 4πr − 256π/r² = 0stationary
  4. 4πr³ = 256π× r²
  5. r³ = 64, so r = 4÷ 4π
  6. h = 128/16 = 8
  7. S = 32π + 64π = 96π
  8. ≈ 302 cm²3 s.f.
  9. radius 4 cm, height 8 cm
Your turn

Question 5 · Merit, in context

  • A poster has 384 cm² of print,
  • margins 3 cm top and bottom,
  • 2 cm each side. Find the paper size
  • that uses the least paper.
Worked example

Q5: paper area in x only

  • Add the margins.
  • Print area gives y.
  1. print: x wide, y high
  2. xy = 384, so y = 384/xprint area
  3. paper: (x + 4) by (y + 6)
  4. A = (x + 4)(y + 6)
  5. = xy + 6x + 4y + 24expand
  6. = 384 + 6x + 1536/x + 244y = 1536/x
  7. = 408 + 6x + 1536/x
Worked example

Q5: the least paper

  • Derivative = 0.
  • Answer for the paper.
  1. A = 408 + 6x + 1536/x
  2. dA/dx = 6 − 1536/x²differentiate
  3. 6 − 1536/x² = 0stationary
  4. x² = 256, so x = 16x > 0
  5. y = 384/16 = 24
  6. 16 cm by 24 cmthe print, not the paper
  7. paper 20 cm by 30 cmx + 4, y + 6
  8. A = 20 × 30 = 600 cm²
Your turn

Question 6 · Excellence

  • Find the point on y = √x nearest
  • to A(2, 0), and its distance.
  • Justify that it is the nearest.
Worked example

Q6: distance as a function of x

  • A point on the curve.
  • Work with D².
  1. point P(x, √x) on the curve
  2. D² = (x − 2)² + (√x − 0)²distance formula
  3. = x² − 4x + 4 + xexpand
  4. = x² − 3x + 4
  5. D least when D² leastD ≥ 0
Worked example

Q6: nearest, and why

  • Excellence: justify the min.
  • Square root at the end.
  1. D² = x² − 3x + 4
  2. d(D²)/dx = 2x − 3differentiate
  3. 2x − 3 = 0, so x = 3/2stationary
  4. d²(D²)/dx² = 2 > 0a minimum
  5. D² = 9/4 − 9/2 + 4 = 7/4
  6. D = √7/2 ≈ 1.323 s.f.
  7. P(1.5, √1.5) ≈ (1.5, 1.22)
  8. nearest distance ≈ 1.32 units
Your turn

Question 7 · Excellence, in context

  • A 30 cm sheet is folded 10 cm from
  • each edge to make a gutter. Each side
  • makes angle θ with the horizontal.
  • Find θ for the greatest cross-section.
Question 7 · Excellence, in context
Worked example

Q7: the area in θ

  • A rectangle and two triangles.
  • Trig for each length.
  1. height = 10sin θside 10 cm
  2. overhang = 10cos θeach side
  3. middle: 10 × 10sin θ = 100sin θ
  4. ends: 2 × (1/2)(10cos θ)(10sin θ)
  5. = 100sin θ cos θ
  6. A = 100sin θ + 100sin θ cos θ
Worked example

Q7: the best angle

  • A quadratic in cos θ.
  • Justify the maximum.
  1. A = 100sin θ + 100sin θ cos θ
  2. dA/dθ = 100(cos θ + cos²θ − sin²θ)
  3. = 100(2cos²θ + cos θ − 1)sin²θ = 1 − cos²θ
  4. = 100(2cos θ − 1)(cos θ + 1)factorise
  5. cos θ = 1/2, so θ = π/3stationary
  6. cos θ = −1: θ = πfolded flat, A = 0
  7. dA/dθ: + before π/3, − aftera maximum
  8. A = 100(√3/2)(1 + 1/2) = 75√3
  9. θ = π/3 (60°): A ≈ 130 cm²
Summary

The method that solved all seven

  • Model, one variable, derivative, answer.
StepHow
Namethe variables, with units
Modelthe quantity to optimise
Constraintdown to one variable
Derivativeset it equal to 0
Rejectvalues that cannot happen
Answerwhat was asked, units

More in AS 91578

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