NCEA Level 2OptimisationAS 91262

Optimisation with calculus: the biggest box

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

Optimisation: the best value

  • Biggest volume, least material, most profit.
  • The best value is at a turning point.
  • So form a function and differentiate it.
Worked example

A box from a 24 cm square card

  • Cut a square of side x from each corner.
  • Fold up the sides: the height is x.
  • Each base side is 24 − 2x.
A box from a 24 cm square card
Worked example

Write the volume in one variable

  • Volume = base² × height.
  • Square the bracket first.
  • Then multiply by x.
  1. height x, base 24 − 2xx off each end
  2. V = x(24 − 2x)²height × base²
  3. (24 − 2x)² = (24 − 2x)(24 − 2x)
  4. = 576 − 48x − 48x + 4x²expand
  5. = 576 − 96x + 4x²collect the x terms
  6. V = x(576 − 96x + 4x²)
  7. = 4x³ − 96x² + 576x
Worked example

Differentiate and solve

  • Take out 12 first.
  • x = 12 leaves no base.
  1. V = 4x³ − 96x² + 576x
  2. V′ = 12x² − 192x + 576differentiate
  3. 12x² − 192x + 576 = 0stationary
  4. 12(x² − 16x + 48) = 0take out 12
  5. 12(x − 4)(x − 12) = 0factorise
  6. x = 4 or x = 12
  7. x = 12: base 24 − 24 = 0no box
  8. x = 4: V = 4(24 − 8)²
  9. = 4 × 256 = 1024 cm³
Key idea

Check the answer makes sense

  • x must be more than 0 and less than 12.
  • An answer outside that is not a box.
  • Say why you reject it.
Worked example

Prove it is a maximum

  • V″ negative: a maximum.
  • Or V′ goes + then −.
  1. V′ = 12x² − 192x + 576
  2. V″ = 24x − 192differentiate again
  3. V″(4) = 24 × 4 − 192
  4. = 96 − 192 = −96 < 0a maximum
  5. V′(3) = 108 − 576 + 576
  6. = 108 > 0rising
  7. V′(5) = 300 − 960 + 576
  8. = −84 < 0falling
  9. x = 4 gives the maximum
Worked example

Answer the question asked

  • Cut 4 cm squares from each corner.
  • The box is 16 cm by 16 cm by 4 cm.
  • Maximum volume: 1024 cm³.
Answer the question asked
Worked example

Another: 60 m of fence against a wall

  • The wall is the fourth side.
  • Two widths, one length.
  1. width = x
  2. x + x + length = 60three sides fenced
  3. length = 60 − 2x
  4. A = x(60 − 2x)width × length
  5. = 60x − 2x²expand
Worked example

Differentiate and solve

  • Set the derivative to zero.
  • Check: a maximum.
  1. A = 60x − 2x²
  2. A′ = 60 − 4xdifferentiate
  3. 60 − 4x = 0stationary
  4. 4x = 60, so x = 15
  5. A″ = −4 < 0a maximum
  6. length = 60 − 30 = 30
  7. A = 15 × 30 = 450 m²
Key idea

Five steps for any optimisation

  • Most marks are lost at step 2 and step 4.
  • One variable only, before differentiating.
  • Reject what the context rules out.
Step
1draw it, name the variable
2write the quantity in one variable
3differentiate, set to zero, solve
4reject answers outside the context
5prove max or min, answer with units
Key idea

Where the marks are

  • Achieved: differentiate the given function.
  • Merit: find the optimum with working.
  • Excellence: form it yourself, justify, check in context.
Summary

What to take away

  • Form it, solve it, reject what cannot be, prove it.
Formone quantity, one variable
Solvederivative equals zero
Domainreject impossible answers
Provesecond derivative or gradient test
Answerthe question asked, with units
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • Question 3 has an answer to reject.
  • Give units.
Question
1A = x(40 − 2x) m². Find the maximum area.
2Profit P = −x² + 50x − 400. Find the maximum profit.
312 cm card, corners x cut: V = x(12 − 2x)². Best x?
4Prove your answer to question 3 is a maximum.
Show answer

1. A′ = 40 − 4x = 0 at x = 10: A = 10 × 20 = 200 m². 2. P′ = −2x + 50 = 0 at x = 25: P = 225. 3. V′ = (12 − 2x)(12 − 6x) is zero at x = 2 or 6; x = 6 leaves no base, so x = 2 cm and V = 128 cm³. 4. V = 4x³ − 48x² + 144x, V″ = 24x − 96, V″(2) = −48 < 0, so a maximum.

Worked example

The answers: questions 1 and 2

  1. 1. A = x(40 − 2x) = 40x − 2x²expand
  2. A′ = 40 − 4x = 0, so x = 10stationary
  3. A = 10(40 − 20) = 10 × 20
  4. = 200 m²
  5. 2. P′ = −2x + 50differentiate
  6. −2x + 50 = 0, so x = 25stationary
  7. P = −625 + 1250 − 400substitute x = 25
  8. = 225
Worked example

The answers: question 3

  1. 3. V = x(12 − 2x)²base 12 − 2x
  2. (12 − 2x)² = 144 − 48x + 4x²square the bracket
  3. V = 4x³ − 48x² + 144xmultiply by x
  4. V′ = 12x² − 96x + 144differentiate
  5. = 12(x² − 8x + 12)take out 12
  6. = 12(x − 2)(x − 6) = 0factorise
  7. x = 6: base 12 − 12 = 0no box
  8. x = 2: V = 2(12 − 4)² = 2 × 64
  9. = 128 cm³
Worked example

The answers: question 4

  1. V′ = 12x² − 96x + 144
  2. 4. V″ = 24x − 96differentiate again
  3. V″(2) = 24 × 2 − 96
  4. = 48 − 96 = −48
  5. −48 < 0, so x = 2 is a maximum

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