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

A mobile plan charges a monthly fee plus a price for each GB of data. Using 4 GB costs £17 and using 10 GB costs £35.

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

   • Gradient = (35 − 17) ÷ (10 − 4) = 3
   • 17 = 3×4 + c, so c = 5

2. (b) Find the cost of using 7.5 GB.

   • \(3\times7.5+5=27.50\)

The weekly profit £P of a concert is modelled by \(P=-3x^2+84x-360\), where £x is the ticket price.

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

   • From the calculator: maximum at x = 14, P = 228

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

   • Solve for x: \(-3x^2+84x-360=0\): x = 5.28 and x = 22.7

An open box is made from a rectangular card 36 cm by 24 cm. A square of side x cm is cut from each corner and the sides are folded up. The volume is \(V=x(36-2x)(24-2x)\) cm³.

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

   • 36 − 2x > 0 and 24 − 2x > 0, and the smaller gives x < 12

2. (b) Find the volume when x = 3.

   • \(3\times30\times18=1620\)

3. (c) Find the greatest volume.

   • From the calculator: maximum at x = 4.71, V = 1830

The braking distance s metres of a car is directly proportional to the square of its speed v m/s: \(s=av^2\). At 20 m/s the braking distance is 24 m.

1. (a) Find a.

   • \(24=a\times20^2\)
   • a = 24 ÷ 400 = 0.06

2. (b) Find the braking distance at 30 m/s.

   • \(0.06\times30^2=54\)

A model is \(f(x)=12+30(0.8)^x\), for \(x\ge0\).

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

   • The curve levels off at c.

2. (b) Find f(0).

   • \(f(0)=12+30=42\)

3. (c) Solve f(x) = 20.

   • \(12+30(0.8)^x=20\)
   • Solve for x: x = 5.92

A tank holds 40 litres of water at t = 0, where t is in minutes, and 100 litres at t = 6. Water flows in at a constant rate. The tank holds at most 300 litres.

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

   • m = (100 − 40) ÷ (6 − 0) = 10
   • c = 40, the amount at t = 0

2. (b) Find the time when there are 250 litres.

   • \(10t+40=250\)
   • t = 21

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

   • \(10t+40=300\) gives t = 26. The tank cannot hold more, and t cannot be negative

A footbridge arch is modelled by \(h=-0.04x^2+1.6x\), where h metres is the height above the water and x metres is the horizontal distance from the left foot of the arch.

1. (a) Find the width of the arch at water level.

   • \(-0.04x^2+1.6x=0\), x = 0 or x = 40

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

   • From the calculator: maximum at x = 20, h = 16

3. (c) Find the width of the arch at a height of 12 m.

   • Solve for x: h = 12 gives x = 10 and x = 30
   • Width = 30 − 10 = 20

The charge level C% of a phone battery t hours after it starts charging is modelled by \(C=100-85a^t\), with 0 < a < 1. After 2 hours the charge is 55%.

1. (a) Find a.

   • \(100-85a^{2}=55\)
   • \(a^{2}=0.5294\), so a = 0.7276

2. (b) Find the charge level after 5 hours.

   • 100 − 85 × 0.7276^5 = 82.7%

3. (c) Find the time for the charge level to reach 95%.

   • \(100-85(0.7276)^t=95\)
   • Solve for t: t = 8.91

The time T hours to paint a hall is inversely proportional to the number n of painters: \(T=kn^{-1}\). Three painters take 14 hours.

1. (a) Find k.

   • \(14=k\times3^{-1}\)
   • k = 42

2. (b) Find the time taken by 7 painters.

   • \(42\div7=6\)

3. (c) Find the least whole number of painters who can finish in under 4 hours.

   • \(42\div n<4\)
   • n > 10.5, so the least whole number is 11

The number N of people (in hundreds) at a festival t hours after it opens is modelled by \(N=-0.5t^3+6t^2\), for \(0\le t\le12\).

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

   • From the calculator: maximum at t = 8, N = 128

2. (b) Find the length of time for which the attendance is at least 10 000.

   • Solve for t: N = 100 gives t = 5.58 and t = 10
   • Between them: 4.42 hours

3. (c) Explain why the model is only used for \(0\le t\le12\).

   • N = 0 when t = 12. After that the model gives a negative attendance, which is impossible.

A cup of tea in a room at 25 °C is measured every 5 minutes. T °C is the temperature after t minutes. The data is: (0, 95), (5, 61), (10, 43), (15, 34), (20, 29). Two models are suggested. A linear model through (0, 95) and (20, 29), and the exponential model \(T=25+70(0.875)^t\).

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

   • m = (29 − 95) ÷ 20 = −3.3, c = 95

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

   • 95 − 3.3 × 40 = −37
   • A cup or a bowl cannot cool below the room.

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

   • 25 + 70 × 0.875^40 = 25.3

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.