NCEA Level 1Optimal solutionsAS 91947

Optimal solutions: 7 exam questions solved step by step

What this lesson covers

  • The best option depends on the quantity being compared, so the comparison has to be set up before the choice is made.
  • Where two cost rules cross is the point at which the better option changes.
Key idea

7 questions, one way of working

  • 1. Name the measure: cheapest, largest.
  • 2. Write a rule for each option.
  • 3. Equal: solve. Better: an inequation.
  • 4. Maximum: halfway between the zeros.
  • 5. Answer in words, with numbers.
Your turn

Question 1 · Achieved, in context

  • Bike hire in Rotorua, for 3 hours.
  • Company A: $15, plus $5 an hour.
  • Company B: $25, plus $3 an hour.
  • Which is cheaper? Justify it.
Worked example

Q1: bike hire for 3 hours

  • Measure: total cost.
  • Fee + rate × hours.
  1. Let h = number of hours
  2. A = 15 + 5hfee + rate × hours
  3. B = 25 + 3h
  4. A = 15 + 5 × 3 = 30h = 3
  5. B = 25 + 3 × 3 = 34h = 3
  6. 30 < 34, a saving of $4
  7. Company A is cheaper, by $4
Your turn

Question 2 · Achieved, in context

  • A gym has two ways to pay.
  • Plan P: $40 to join, then $8 a visit.
  • Plan Q: $12 a visit, nothing to join.
  • When do both cost the same?
Worked example

Q2: the break-even point

  • Same cost: set equal.
  • Check in both rules.
  1. P = 40 + 8nn = number of visits
  2. Q = 12nno joining fee
  3. 40 + 8n = 12nP = Q
  4. 40 = 4n−8n both sides
  5. n = 10÷ 4
  6. P = 40 + 80 = 120
  7. Q = 12 × 10 = 120
  8. 10 visits, both cost $120
Your turn

Question 3 · Merit, in context

  • Bike hire: A = 15 + 5h, B = 25 + 3h
  • Hire is charged in whole hours.
  • For which hires is B cheaper?
  • Answer in hours, not just in h.
Worked example

Q3: when is B cheaper?

  • Cheaper: less than.
  • Whole hours only.
  1. 25 + 3h < 15 + 5hB < A
  2. 25 < 15 + 2h−3h both sides
  3. 10 < 2h−15 both sides
  4. 5 < h, so h > 5÷ 2, sign stays
  5. h = 5: A = 40, B = 40break-even
  6. h = 6: A = 45, B = 43B wins
  7. 6 hours or more: B is cheaper
Your turn

Question 4 · Merit, in context

  • A vege garden against a wall.
  • 20 m of fence for the other 3 sides.
  • Width x m, so A = x(20 − 2x).
  • Find the largest possible area.
Question 4 · Merit, in context
Worked example

Q4: the largest garden

  • Opens downward: a max.
  • Max is halfway.
  1. length = 20 − 2xtwo widths use 2x
  2. A = x(20 − 2x) = −2x² + 20xx² negative: a max
  3. A = 0: x = 0 or 20 − 2x = 0
  4. zeros: x = 0 and x = 10
  5. x = (0 + 10) ÷ 2 = 5halfway
  6. length = 20 − 2 × 5 = 10
  7. A = 5 × 10 = 50
  8. x = 4 or 6: A = 48 < 50
  9. 50 m², 5 m wide by 10 m long
Your turn

Question 5 · Merit, in context

  • A school quiz night sells tickets.
  • At $p each, it sells 200 − 10p.
  • Income: R = p(200 − 10p) dollars.
  • Which price gives the most income?
Worked example

Q5: the best ticket price

  • Measure: income.
  • Halfway between zeros.
  1. R = −10p² + 200popens downward
  2. R = 0: p = 0 or 200 − 10p = 0
  3. zeros: p = 0 and p = 20
  4. p = (0 + 20) ÷ 2 = 10halfway
  5. tickets = 200 − 10 × 10 = 100
  6. R = 10 × 100 = 1000
  7. p = 9: 9 × 110 = 990
  8. p = 11: 11 × 90 = 990
  9. $10 a ticket, income $1000
Your turn

Question 6 · Excellence

  • Two sheep pens, side by side,
  • against a woolshed wall.
  • 24 m of fence, one across the middle.
  • Find the pens with the most area.
Question 6 · Excellence
Worked example

Q6: two pens, one fence

  • The divider is a width.
  • Then halfway again.
  1. Let x = width of the pens (m)
  2. length = 24 − 3x3 widths use 3x
  3. A = x(24 − 3x) = −3x² + 24xopens downward
  4. zeros: x = 0 and x = 824 − 3x = 0
  5. x = (0 + 8) ÷ 2 = 4halfway
  6. length = 24 − 3 × 4 = 12
  7. A = 4 × 12 = 48
  8. x = 3 or 5: A = 45 < 48
  9. 4 m by 12 m, 48 m² in total
Your turn

Question 7 · Excellence

  • C: $f fixed, plus $4 an hour.
  • D: $7 an hour, no fixed fee.
  • Show they break even at h = f/3.
  • Find f for 4 hours; when is C cheaper?
Worked example

Q7: break-even with a letter

  • Same method, a letter.
  • Solve for h.
  1. C = f + 4h
  2. D = 7h
  3. f + 4h = 7hC = D
  4. f = 3h−4h both sides
  5. h = f/3÷ 3
  6. h = 4: f = 3 × 4 = 12
  7. C = 12 + 16 = 28, D = 28
Worked example

Q7: when is C cheaper?

  • Cheaper: less than.
  • Whole hours only.
  1. C = 12 + 4h, D = 7h
  2. 12 + 4h < 7hC < D
  3. 12 < 3h−4h both sides
  4. 4 < h, so h > 4÷ 3
  5. h = 5: C = 32, D = 35
  6. f = $12; C cheaper from 5 hours
Summary

The method that solved all seven

  • Name the measure, compare, and justify in words.
StepHow
The measurename what best means first
Two rulesfixed fee + rate × amount
Break-evenset the rules equal, solve
Better whenan inequation, then check
Whole unitsh > 5 means 6 hours or more
Most areahalfway between the zeros
Justify itchoice, numbers and reason

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