L48 — Optimisation and Decision Making — Solutions

← Back to Problem Solving and Modelling

1. Q1 — Maximum area with fixed perimeter

   Fluency
   • (a) Perimeter = 2x + 2l = 60 ⇒ l = (60−2x)/2 = 30−x.
   • (b) A(x) = x(30−x) = −x² + 30x.
   • (c) Vertex: x = −30/(2×−1) = 15 cm.
   • (d) 15 cm × 15 cm (a square). Maximum area = 15² = 225 cm².

2. Q2 — Vertex of a profit function

   Fluency
   • (a) n = −80/(2×−2) = 20 items.
   • (b) P(20) = −2(400)+80(20)−300 = −800+1600−300 = $500.
   • (c) −2n²+80n−300 < 0 ⇒ n²−40n+150 > 0 ⇒ (n−5)(n−30) > 0. Loss when n < 5 or n > 30.
   • (d) n must be a positive whole number: n ≥ 1, n ∈ ℤ. Also, production usually has a practical upper limit.

3. Q3 — Maximum revenue

   Fluency
   • (a) Price = (8 + p). Quantity = (200 − 10p).
   • (b) R(p) = (8+p)(200−10p) = 1600 − 80p + 200p − 10p² = −10p² + 120p + 1600.
   • (c) p = −120/(2×−10) = 6. Price = 8 + 6 = $14 per bag.
   • (d) R(6) = −10(36)+120(6)+1600 = −360+720+1600 = $1 960 per day.

4. Q4 — Fence problem — one wall given

   Fluency
   • (a) Two widths + one length = 120: 2x + L = 120 ⇒ L = 120 − 2x.
   • (b) A(x) = x(120−2x) = −2x²+120x. Vertex: x = 30. A(30) = 30 × 60 = 1800 m².
   • (c) Width = 30 m, Length = 60 m.
   • (d) x > 0 and L > 0: 120−2x > 0 ⇒ x < 60. Domain: 0 < x < 60.

5. Q5 — Ticket pricing optimisation

   Understanding
   • (a) Ticket price = (25 + 5n). Tickets sold = (1000 − 50n).
   • (b) R(n) = (25+5n)(1000−50n) = 25000 − 1250n + 5000n − 250n² = −250n² + 3750n + 25000.
   • (c) n = −3750/(2×−250) = 7.5. So n = 7 or 8 (whole increases). Check: R(7)=250(49)+3750(7)+25000=$49125+25000... Let's compute properly. At n=7.5, price = 25+37.5 = $62.50 (round to n=7: $60 or n=8: $65).
   • (d) At n=20, tickets sold = 0. So domain: 0 ≤ n ≤ 20. Vertex n=7.5 is inside domain ✓. Maximum at n=7 (R=$49 350) or n=8 (R=$49 250), so n=7 ⇒ $60 ticket gives max revenue.

6. Q6 — Minimum material for a box

   Understanding
   • (a) x²h = 250 ⇒ h = 250/x².
   • (b) S(x) = x² + 4xh = x² + 4x(250/x²) = x² + 1000/x.
   • (c) S(5) = 25+200 = 225. S(6) = 36+166.7 ≈ 202.7. S(7) = 49+142.9 ≈ 191.9. S is smallest at x = 7 (continues decreasing then increases — minimum ≈ x=6.3 by calculus, so x=6 or 7 are close).
   • (d) Minimising surface area minimises the amount of material needed — reducing manufacturing cost for a given volume of product.

7. Q7 — Comparing options

   Understanding
   • (a) Square: side = 24/4 = 6 m. Area = 36 m².
   • (b) Rectangle: 2w + 2(2w) = 24 ⇒ 6w = 24 ⇒ w = 4 m, l = 8 m. Area = 32 m².
   • (c) The square has larger area (36 m² > 32 m²).
   • (d) For a fixed perimeter, the square always gives the largest area among rectangles. This is because the area is maximised when the dimensions are equal (vertex of the quadratic).

8. Q8 — Endpoint check

   Understanding
   • (a) Vertex: x = −8/(2×−1) = 4. P(4) = −16+32−7 = 9.
   • (b) x = 4 is within [1, 5] ⇒ yes, vertex is in the domain.
   • (c) P(1) = −1+8−7 = 0. P(4) = 9. P(5) = −25+40−7 = 8. Maximum is P(4) = 9 (at x = 4 thousand units).
   • (d) If the vertex is outside the domain, the maximum must occur at one of the endpoints. Without checking endpoints, you might report a maximum that isn't achievable under the given constraints.

9. Q9 — Optimisation in context

   Problem Solving
   • (a) R(x) = x(50−2x) = −2x² + 50x ($thousands).
   • (b) P(x) = −2x²+50x − 10x − 30 = −2x² + 40x − 30.
   • (c) Vertex: x = −40/(2×−2) = 10 thousand units.
   • (d) P(10) = −2(100)+400−30 = −200+370 = $170 thousand profit. Price = 50−2(10) = $30 per unit.

10. Q10 — Extended optimisation problem

    Problem Solving
    • (a) Track perimeter = 2L + 2πr = 400. 2L = 400−2πr ⇒ L = 200 − πr. ✓
    • (b) Rectangle area = 2r × L. Two semicircles = πr². Total A = 2rL + πr² = 2r(200−πr) + πr² = 400r − 2πr² + πr² = 400r − πr².
    • (c) A(r) = −πr² + 400r. Vertex: r = −400/(2×−π) = 200/π ≈ 63.7 m.
    • (d) A(200/π) = 400(200/π) − π(200/π)² = 80000/π − 40000/π = 40000/π ≈ 12 732 m². When L=0, the shape is a circle — the optimal enclosure for a given perimeter.