Solutions — Optimisation Problems

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1. Show Solution
   f(x) = −x² + 6x + 1 on [0, 5]
   
   Step 1 — Find critical point:
   f′(x) = −2x + 6 = 0 ⇒ x = 3
   
   Step 2 — Confirm maximum:
   f″(x) = −2 < 0 ⇒ x = 3 is a local maximum
   
   Step 3 — Evaluate at critical point and endpoints:
   f(0) = 0 + 0 + 1 = 1
   f(3) = −9 + 18 + 1 = 10
   f(5) = −25 + 30 + 1 = 6
   
   Maximum value is 10, occurring at x = 3.
2. Show Solution
   Define variables:
   Let width = x m (perpendicular to wall, two sides needed), length = y m (parallel to wall, one side).
   Objective: maximise A = xy
   
   Constraint:
   2x + y = 100 ⇒ y = 100 − 2x
   A(x) = x(100 − 2x) = 100x − 2x²
   Domain: x > 0 and y = 100 − 2x > 0 ⇒ 0 < x < 50
   
   Differentiate:
   A′(x) = 100 − 4x = 0 ⇒ x = 25
   
   Confirm maximum:
   A″(x) = −4 < 0 ⇒ maximum at x = 25
   y = 100 − 50 = 50
   
   Width = 25 m, length = 50 m. Maximum area = 25 × 50 = 1250 m².
3. Show Solution
   Let the two numbers be x and y where x + y = 20, so y = 20 − x.
   Objective: maximise P = xy
   P(x) = x(20 − x) = 20x − x²; domain: 0 < x < 20
   
   Differentiate:
   P′(x) = 20 − 2x = 0 ⇒ x = 10
   
   Confirm maximum:
   P″(x) = −2 < 0 ⇒ maximum at x = 10
   y = 20 − 10 = 10
   
   The two numbers are both 10. Maximum product = 10 × 10 = 100.
   Note: for a fixed sum, the product is maximised when both numbers are equal — a useful result to remember.
4. Show Solution
   S(x) = x + 36/x = x + 36x⁻¹; domain: x > 0
   
   Differentiate:
   S′(x) = 1 − 36x⁻² = 1 − 36/x²
   S′(x) = 0 ⇒ 1 = 36/x² ⇒ x² = 36 ⇒ x = 6 (since x > 0)
   
   Confirm minimum:
   S″(x) = 72x⁻³ = 72/x³
   S″(6) = 72/216 = 1/3 > 0 ⇒ minimum confirmed
   
   Minimum value: S(6) = 6 + 36/6 = 6 + 6 = 12.
   This minimum occurs at x = 6.
5. Show Solution
   Let base side = x m and height = h m.
   
   Constraint (volume):
   x²h = 32 ⇒ h = 32/x²
   
   Objective function (surface area, open top):
   S = base + 4 sides = x² + 4xh = x² + 4x · (32/x²) = x² + 128/x
   Domain: x > 0
   
   Differentiate:
   S′(x) = 2x − 128/x²
   S′(x) = 0 ⇒ 2x = 128/x² ⇒ 2x³ = 128 ⇒ x³ = 64 ⇒ x = 4
   
   Confirm minimum:
   S″(x) = 2 + 256/x³
   S″(4) = 2 + 256/64 = 2 + 4 = 6 > 0 ⇒ minimum confirmed
   
   h = 32/4² = 32/16 = 2
   
   Base side = 4 m, height = 2 m. Minimum surface area = 4² + 128/4 = 16 + 32 = 48 m².
6. Show Solution
   Let radius = r cm and height = h cm.
   
   Constraint (volume):
   πr²h = 250π ⇒ h = 250/r²
   
   Objective function (surface area, no lid):
   S = πr² + 2πrh = πr² + 2πr · (250/r²) = πr² + 500π/r
   Domain: r > 0
   
   Differentiate:
   S′(r) = 2πr − 500π/r²
   S′(r) = 0 ⇒ 2πr = 500π/r² ⇒ 2r³ = 500 ⇒ r³ = 250
   r = 250^1/3 = 5 · 2^1/3 ≈ 6.30 cm
   
   Confirm minimum:
   S″(r) = 2π + 1000π/r³ > 0 for all r > 0 ⇒ minimum confirmed
   
   h = 250/r² = 250/250^2/3 = 250^1/3 = r ≈ 6.30 cm
   (Note: the optimal cylinder has h = r, i.e., height equals radius.)
   
   Radius = 5 · 2^1/3 ≈ 6.30 cm, height ≈ 6.30 cm for minimum surface area.
7. Show Solution
   P(x) = −2x² + 120x − 800; domain: x ≥ 0
   
   Differentiate:
   P′(x) = −4x + 120
   P′(x) = 0 ⇒ −4x + 120 = 0 ⇒ x = 30
   
   Confirm maximum:
   P″(x) = −4 < 0 ⇒ maximum at x = 30
   
   Maximum profit:
   P(30) = −2(900) + 120(30) − 800
   = −1800 + 3600 − 800
   = $1000
   
   Profit is maximised when 30 units are sold. Maximum profit = $1000.
8. Show Solution
   x(t) = t³ − 6t² + 9t on [0, 4]
   
   Find critical points:
   x′(t) = 3t² − 12t + 9 = 3(t² − 4t + 3) = 3(t − 1)(t − 3)
   x′(t) = 0 ⇒ t = 1 or t = 3 (both in domain)
   
   Evaluate x at all candidates:
   x(0) = 0
   x(1) = 1 − 6 + 9 = 4
   x(3) = 27 − 54 + 27 = 0
   x(4) = 64 − 96 + 36 = 4
   
   Maximum value = 4 (occurring at t = 1 and t = 4).
   Minimum value = 0 (occurring at t = 0 and t = 3).
   
   Maximum displacement from origin = 4 m.
9. Show Solution
   Let x cm be used for the circle, so (10 − x) cm for the square. Domain: 0 ≤ x ≤ 10.
   
   Circle: circumference = x ⇒ 2πr = x ⇒ r = x/(2π)
   Area of circle = πr² = π · x²/(4π²) = x²/(4π)
   
   Square: perimeter = 10 − x ⇒ side = (10 − x)/4
   Area of square = (10 − x)²/16
   
   Total area:
   A(x) = x²/(4π) + (10 − x)²/16
   
   Differentiate:
   A′(x) = 2x/(4π) + 2(10 − x)(−1)/16 = x/(2π) − (10 − x)/8
   A′(x) = 0: x/(2π) = (10 − x)/8
   8x = 2π(10 − x)
   8x = 20π − 2πx
   x(8 + 2π) = 20π
   x = 20π/(8 + 2π) = 10π/(4 + π) ≈ 4.40 cm
   
   Confirm minimum:
   A″(x) = 1/(2π) + 1/8 > 0 ⇒ minimum
   
   Cut the wire at 10π/(4+π) ≈ 4.40 cm from one end. Use ≈ 4.40 cm for the circle and ≈ 5.60 cm for the square to minimise total enclosed area.
   (Note: to maximise total area, use all wire for the circle — endpoint x = 10 gives area 100/(4π) ≈ 7.96; all for square gives (10/4)² = 6.25. So maximum is at x = 10, entirely a circle.)
10. Show Solution
    Place the semicircle with its diameter on the x-axis, centre at the origin. The semicircle has equation x² + y² = 25 with y ≥ 0.
    
    Set up:
    Let the rectangle have half-width x (so full width = 2x) and height y, with the top corners on the semicircle.
    Constraint: x² + y² = 25 ⇒ y = √(25 − x²)
    Objective: maximise A = 2xy = 2x√(25 − x²); domain: 0 < x < 5
    
    Differentiate (product rule):
    A′(x) = 2√(25 − x²) + 2x · (−x)/√(25 − x²)
    = [2(25 − x²) − 2x²] / √(25 − x²)
    = (50 − 4x²) / √(25 − x²)
    
    Set A′(x) = 0:
    50 − 4x² = 0 ⇒ x² = 12.5 ⇒ x = 5/√2 = 5√2/2
    
    y = √(25 − 12.5) = √(12.5) = 5/√2 = 5√2/2
    
    Dimensions and maximum area:
    Width = 2x = 5√2, Height = y = 5√2/2
    A = 2 · (5√2/2) · (5√2/2) = 5√2 · 5√2/2 = 50/2 = 25
    
    Largest rectangle has width 5√2 and height 5√2/2. Maximum area = 25 square units.