Optimisation in context practice questions.

A company's profit (in thousands of dollars) is modelled by P(x) = -2x² + 48x - 150, where x is the number of units sold (in hundreds). Determine the number of units (in hundreds) that maximises the profit.

• 12
• 24
• 6
• 48

Answer: 12

At a maximum P'(x) = -4x + 48 = 0, so x = 12. The vertex of the downward parabola gives the maximising value.

A rectangular field is to be enclosed using 200 m of fencing on all four sides. Determine the maximum possible area of the field.

• 2000 m²
• 2500 m²
• 10000 m²
• 5000 m²

Answer: 2500 m²

For a fixed perimeter the maximum area is a square: each side is 200/4 = 50 m, so area = 50 × 50 = 2500 m².

The height of a projectile is h(t) = 20t - 5t² metres, where t is time in seconds. Determine the maximum height reached.

• 40 m
• 10 m
• 20 m
• 15 m

Answer: 20 m

h'(t) = 20 - 10t = 0 gives t = 2 s. Then h(2) = 40 - 20 = 20 m.

The function f(x) = x³ - 3x² - 9x + 5 has two stationary points. Determine the x-coordinate of the local minimum.

• -1
• 3
• 1
• -3

Answer: 3

f'(x) = 3x² - 6x - 9 = 3(x-3)(x+1) = 0 gives x = 3 or x = -1. Since f''(x) = 6x - 6, at x = 3 f'' = 12 > 0, so x = 3 is the local minimum.

The revenue from selling x items is R(x) = 90x - 3x² dollars. Determine the number of items that maximises revenue.

• 30
• 45
• 15
• 90

Answer: 15

R'(x) = 90 - 6x = 0 gives x = 15. Since R'' = -6 < 0 this is a maximum.

The number of stationary points of the cubic f(x) = x³ + 6x² + 12x + 7 is determined by its derivative. Determine how many stationary points f has.

• 2
• 1
• 0
• 3

Answer: 1

f'(x) = 3x² + 12x + 12 = 3(x + 2)². This is zero only at x = -2 (a repeated root), so there is exactly one stationary point (a point of inflexion).