Optimisation in context Problem set A One question at a time

The power output of a solar panel, P watts, at t hours after midnight is modelled by \(P=-0.8t^2+19.2t-90\), for \(7\le t\le 17\). Find

1. (a) the time when the output is greatest;

   • Graph \(y=-0.8x^2+19.2x-90\) for \(7\le x\le 17\)
   • From the calculator: maximum at (12, 25.2)
   • t = 12 hours

2. (b) the greatest output.

   • The maximum value is the y-coordinate of that point
   • 25.2 watts

The running cost of a lorry is \(C=0.01v^2+\frac{160}{v}\) euros per kilometre, where v km/h is its speed. Find

1. (a) the speed that makes the cost smallest;

   • Graph \(y=0.01x^2+\frac{160}{x}\) for \(x>0\)
   • From the calculator: minimum at (20, 12)
   • v = 20 km/h

2. (b) the smallest cost per kilometre.

   • The minimum value is the y-coordinate of that point
   • 12 euros per km

A festival sells a fraction s of its tickets in advance. Its revenue, in thousands of euros, is modelled by \(R=30s(1-s)^2\), for \(0\le s\le 1\). Find

1. (a) the fraction that gives the greatest revenue;

   • Graph \(y=30x(1-x)^2\) for \(0\le x\le 1\)
   • From the calculator: maximum at (0.333, 4.44)
   • s = 0.333

2. (b) the greatest revenue.

   • The maximum value is the y-coordinate of that point
   • 4.44 thousand euros

A closed box has a rectangular base measuring \(2x\) cm by x cm, and a height of h cm. Its total surface area is 540 cm².

1. (a) Write the height h in terms of x.

   • Surface area: \(4x^2+6xh=540\)
   • \(h=\frac{540-4x^2}{6x}\)

2. (b) Write the volume V in terms of x only.

   • \(V=2x^2h\)
   • \(V=2x^2\cdot\frac{540-4x^2}{6x}=180x-\frac{4}{3}x^3\)

3. (c) Find the greatest possible volume, and the value of x that gives it.

   • Graph \(y=180x-\frac{4}{3}x^3\) for x > 0
   • From the calculator: maximum at (6.71, 805)
   • V = 805 cm³, when x = 6.71

A farmer has 160 m of fencing to make a rectangular pen against a straight wall. The wall forms one side, so only three sides need fencing. The two sides at right angles to the wall are x metres long.

1. (a) Write the area A in terms of x.

   • \(y=160-2x\) (the fence has \(2x+y=160\))
   • \(A=xy=x(160-2x)=160x-2x^2\)

2. (b) Find the greatest possible area, and the value of x that gives it.

   • Graph \(y=160x-2x^2\) for \(0<x<80\)
   • From the calculator: maximum at (40, 3200)
   • A = 3200 m², when x = 40 m

A rectangular sheet of card measures 60 cm by 36 cm. A square of side x cm is cut from each corner and the sides are folded up to make an open gift box.

1. (a) Write the length and the width of the base in terms of x.

   • Two squares of side x come off each length, so \(60-2x\)
   • Two come off each width, so \(36-2x\)

2. (b) Write the volume V in terms of x, and state the values x can take.

   • \(V=x(60-2x)(36-2x)\)
   • The width must be positive: \(36-2x>0\), so \(0<x<18\)

3. (c) Find the greatest possible volume, and the value of x that gives it.

   • Graph \(y=x(60-2x)(36-2x)\) for \(0<x<18\)
   • From the calculator: maximum at (7.28, 7090)
   • V = 7090 cm³, when x = 7.28

The flow rate of a river, F cubic metres per second, at a distance d hundred metres downstream of a weir is modelled by \(F=-0.01d^3+0.3d^2\), for \(0\le d\le 30\). Find

1. (a) the distance where the flow is greatest;

   • Graph \(y=-0.01x^3+0.3x^2\) for \(0\le x\le 30\)
   • From the calculator: maximum at (20, 40)
   • d = 20 hundred metres

2. (b) the greatest flow rate.

   • The maximum value is the y-coordinate of that point
   • 40 m³/s

A storage tank has no lid. It has a square base of side x metres and a height of h metres, and its volume is 32 m³. The tank is made from steel sheet.

1. (a) Write the height h in terms of x.

   • \(V=x^2h=32\)
   • \(h=\frac{32}{x^2}\)

2. (b) Write the surface area S in terms of x only.

   • The base has area \(x^2\) and the four sides have area \(4xh\)
   • \(S=x^2+4x\cdot\frac{32}{x^2}=x^2+\frac{128}{x}\)

3. (c) Find the smallest possible surface area, and the value of x that gives it.

   • Graph \(y=x^2+\frac{128}{x}\) for x > 0
   • From the calculator: minimum at (4, 48)
   • S = 48 m², when x = 4

A bakery sells 250 loaves a day at €3 each. For every 10 cent rise in the price, 6 fewer loaves are sold each day. Each loaf costs €1 to make. Let x be the number of 10 cent rises.

1. (a) Write an expression for the price and for the number sold after x rises.

   • Price: \(3+0.1x\) euros
   • Number sold: \(250-6x\)

2. (b) Write an expression for the daily profit P in euros.

   • Profit per item: \(3+0.1x-1=2+0.1x\)
   • \(P=(2+0.1x)(250-6x)\)

3. (c) Find the greatest profit when x must be a whole number, and the price that gives it.

   • Graph \(y=(2+0.1x)(250-6x)\). From the calculator: maximum at x = 10.83
   • x must be a whole number. 10: P = 570.00; 11: P = 570.40
   • x = 11 gives the larger profit. Price \(=3+0.1(11)=4.10\) euros; profit €570.40

A school running track has two straight sides of length L metres. At each end there is a semicircle of diameter w metres. The perimeter of the track is 300 m.

1. (a) Write L in terms of w.

   • \(2L+\pi w=300\)
   • \(L=\frac{300-\pi w}{2}\)

2. (b) Write the area A of the rectangle in terms of w.

   • \(A=Lw=\frac{w(300-\pi w)}{2}\)

3. (c) Find the greatest possible area of the rectangle, and the value of w that gives it.

   • Graph \(y=\frac{x(300-\pi x)}{2}\) for \(0<x<300/\pi\)
   • From the calculator: maximum at (47.7, 3580)
   • A = 3580 m², when w = 47.7 m

A closed cylindrical tin has radius r cm and height h cm. Its total surface area is 600 cm².

1. (a) Write V in terms of r only.

   • \(2\pi r^2+2\pi rh=600\), so \(h=\frac{600-2\pi r^2}{2\pi r}\)
   • \(V=\pi r^2h=300r-\pi r^3\)

2. (b) Find the greatest possible volume, and the value of r that gives it.

   • Graph \(y=300x-\pi x^3\) for x > 0
   • From the calculator: maximum at (5.64, 1130)
   • V = 1130 cm³, when r = 5.64 cm

A rectangular plot measuring x metres by y metres has an area of 800 m². It is fenced. Fencing along the two sides of length x costs €15 per metre, and along the two sides of length y costs €6 per metre.

1. (a) Write the cost C in terms of x only.

   • \(xy=800\), so \(y=\frac{800}{x}\)
   • \(C=30x+12y=30x+\frac{9600}{x}\)

2. (b) Find the least possible cost, and the dimensions of the rectangle.

   • Graph \(y=30x+\frac{9600}{x}\) for x > 0
   • From the calculator: minimum at (17.9, 1070)
   • Least cost €1073.31; x = 17.9 m and y = 44.7 m