Linear, quadratic, cubic, variation and exponential models Problem set B Worked solutions

A car-hire firm charges a fixed fee plus a price for each day. Hiring for 3 days costs £96 and hiring for 8 days costs £226.

1. (a) Find a model of the form \(C=md+c\).

   • Gradient = (226 − 96) ÷ (8 − 3) = 26
   • 96 = 26×3 + c, so c = 18

2. (b) Find the cost of hiring for 6 days.

   • \(26\times6+18=174.00\)

The daily profit £P of a bakery stall is modelled by \(P=-2x^2+52x-240\), where £x is the price of a loaf box.

1. (a) Find the price that gives the greatest profit, and that profit.

   • From the calculator: maximum at x = 13, P = 98

2. (b) Find the two prices at which the profit is zero.

   • Solve for x: \(-2x^2+52x-240=0\): x = 6 and x = 20

A gardener makes an open planter from a rectangular sheet of plywood 50 cm by 30 cm, by cutting a square of side x cm from each corner and folding up the sides. The volume is \(V=x(50-2x)(30-2x)\) cm³.

1. (a) Write down a reasonable domain for x.

   • 50 − 2x > 0 and 30 − 2x > 0, and the smaller gives x < 15

2. (b) Find the volume when x = 4.

   • \(4\times42\times22=3696\)

3. (c) Find the greatest volume.

   • From the calculator: maximum at x = 6.07, V = 4100

The power P kilowatts of a wind turbine is modelled by \(P=av^3\), where v m/s is the wind speed. At 6 m/s the power is 54 kW.

1. (a) Find a.

   • \(54=a\times6^3\)
   • a = 54 ÷ 216 = 0.25

2. (b) Find the power at 10 m/s.

   • \(0.25\times10^3=250\)

A model is \(f(x)=5+24(0.85)^x\), for \(x\ge0\).

1. (a) Write down the equation of the horizontal asymptote.

   • The curve levels off at c.

2. (b) Find f(0).

   • \(f(0)=5+24=29\)

3. (c) Solve f(x) = 14.

   • \(5+24(0.85)^x=14\)
   • Solve for x: x = 6.04

A bath holds 18 litres of water at t = 0, where t is in minutes, and 66 litres at t = 4. Water flows in at a constant rate. The bath holds at most 150 litres.

1. (a) Find a model \(V=mt+c\).

   • m = (66 − 18) ÷ (4 − 0) = 12
   • c = 18, the amount at t = 0

2. (b) Find the time when there are 120 litres.

   • \(12t+18=120\)
   • t = 8.5

3. (c) Write down a reasonable domain for the model.

   • \(12t+18=150\) gives t = 11. The tank cannot hold more, and t cannot be negative

A tunnel entrance is modelled by \(h=-0.05x^2+1.5x\), where h metres is the height above the road and x metres is the horizontal distance from the left wall.

1. (a) Find the width of the tunnel at road level.

   • \(-0.05x^2+1.5x=0\), x = 0 or x = 30

2. (b) Find the greatest height of the tunnel.

   • From the calculator: maximum at x = 15, h = 11.25

3. (c) Find the width of the tunnel at a height of 8 m.

   • Solve for x: h = 8 gives x = 6.94 and x = 23.06
   • Width = 23.06 − 6.94 = 16.1

A hot metal rod cools in a room. Its temperature T °C after t minutes is modelled by \(T=21+79a^t\), with 0 < a < 1. After 4 minutes the temperature is 50 °C.

1. (a) Find a.

   • \(21+79a^{4}=50\)
   • \(a^{4}=0.3671\), so a = 0.7784

2. (b) Find the temperature after 10 minutes.

   • 21 + 79 × 0.7784^10 = 27.4 °C

3. (c) Find the time for the temperature to fall to 25 °C.

   • \(21+79(0.7784)^t=25\)
   • Solve for t: t = 11.9

The time D days to dig a trench is inversely proportional to the number n of workers: \(D=kn^{-1}\). Five workers take 18 days.

1. (a) Find k.

   • \(18=k\times5^{-1}\)
   • k = 90

2. (b) Find the time taken by 6 workers.

   • \(90\div6=15\)

3. (c) Find the least whole number of workers who can finish in under 8 days.

   • \(90\div n<8\)
   • n > 11.25, so the least whole number is 12

The volume V (in millions of litres) of water in a reservoir t weeks after the rains stop is modelled by \(V=-t^3+9t^2\), for \(0\le t\le9\).

1. (a) Find when the volume is greatest, and the greatest volume.

   • From the calculator: maximum at t = 6, V = 108

2. (b) Find the length of time for which the volume is at least 80 million litres.

   • Solve for t: V = 80 gives t = 4 and t = 7.62
   • Between them: 3.62 weeks

3. (c) Explain why the model is only used for \(0\le t\le9\).

   • V = 0 when t = 9. After that the model gives a negative volume, which is impossible.

A bowl of soup in a room at 18 °C is measured every 5 minutes. T °C is the temperature after t minutes. The data is: (0, 90), (5, 54), (10, 36), (15, 27), (20, 22). Two models are suggested. A linear model through (0, 90) and (20, 22), and the exponential model \(T=18+72(0.87)^t\).

1. (a) Find the linear model \(T=c+mt\).

   • m = (22 − 90) ÷ 20 = −3.4, c = 90

2. (b) Find the temperature at t = 40 minutes predicted by the linear model, and say why it is unreasonable.

   • 90 − 3.4 × 40 = −46
   • A cup or a bowl cannot cool below the room.

3. (c) Find the temperature at t = 40 minutes predicted by the exponential model.

   • 18 + 72 × 0.87^40 = 18.3

4. (d) Which model is better for long times? Give a reason.

   • The data levels off, and the exponential model has an asymptote at the room temperature. The linear model has no asymptote.