Optimisation in context Problem set B One question at a time

The number of customers per hour entering a café, S, at t hours after midnight is modelled by \(S=-1.5t^2+33t-120\), for \(6\le t\le 16\). Find

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

   • Graph \(y=-1.5x^2+33x-120\) for \(6\le x\le 16\)
   • From the calculator: maximum at (11, 61.5)
   • t = 11 hours

2. (b) the greatest number of customers per hour.

   • The maximum value is the y-coordinate of that point
   • 61.5 customers per hour

The fuel cost of a ship is \(C=0.5v^2+\frac{1728}{v}\) dollars per hour, where v knots is its speed. Find

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

   • Graph \(y=0.5x^2+\frac{1728}{x}\) for \(x>0\)
   • From the calculator: minimum at (12, 216)
   • v = 12 knots

2. (b) the smallest cost per hour.

   • The minimum value is the y-coordinate of that point
   • 216 dollars per hour

A bakery makes a fraction f of its dough into rolls. Its profit, in hundreds of pounds, is modelled by \(P=12f(1-f)^2\), for \(0\le f\le 1\). Find

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

   • Graph \(y=12x(1-x)^2\) for \(0\le x\le 1\)
   • From the calculator: maximum at (0.333, 1.78)
   • f = 0.333

2. (b) the greatest profit.

   • The maximum value is the y-coordinate of that point
   • 1.78 hundred pounds

A closed storage crate has a rectangular base measuring \(3x\) cm by x cm, and a height of h cm. Its total surface area is 864 cm².

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

   • Surface area: \(6x^2+8xh=864\)
   • \(h=\frac{864-6x^2}{8x}\)

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

   • \(V=3x^2h\)
   • \(V=3x^2\cdot\frac{864-6x^2}{8x}=324x-2.25x^3\)

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

   • Graph \(y=324x-2.25x^3\) for x > 0
   • From the calculator: maximum at (6.93, 1500)
   • V = 1500 cm³, when x = 6.93

A gardener has 84 m of edging to enclose a rectangular flower bed against a straight garden wall. The wall forms one side, so only three sides need edging. The two sides at right angles to the wall are x metres long.

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

   • \(y=84-2x\) (the fence has \(2x+y=84\))
   • \(A=xy=x(84-2x)=84x-2x^2\)

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

   • Graph \(y=84x-2x^2\) for \(0<x<42\)
   • From the calculator: maximum at (21, 882)
   • A = 882 m², when x = 21 m

A rectangular sheet of tin measures 54 cm by 30 cm. A square of side x cm is cut from each corner and the sides are folded up to make an open planter.

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 \(54-2x\)
   • Two come off each width, so \(30-2x\)

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

   • \(V=x(54-2x)(30-2x)\)
   • The width must be positive: \(30-2x>0\), so \(0<x<15\)

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

   • Graph \(y=x(54-2x)(30-2x)\) for \(0<x<15\)
   • From the calculator: maximum at (6.19, 4540)
   • V = 4540 cm³, when x = 6.19

The speed of a rocket sled, v metres per second, at t seconds after the start is modelled by \(v=-0.02t^3+0.45t^2\), for \(0\le t\le 22.5\). Find

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

   • Graph \(y=-0.02x^3+0.45x^2\) for \(0\le x\le 22.5\)
   • From the calculator: maximum at (15, 33.8)
   • t = 15 seconds

2. (b) the greatest speed.

   • The maximum value is the y-coordinate of that point
   • 33.8 m/s

A fish crate has no lid. It has a square base of side x cm and a height of h cm, and its volume is 4000 cm³. The crate is made from plastic sheet.

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

   • \(V=x^2h=4000\)
   • \(h=\frac{4000}{x^2}\)

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

   • The base has area \(x^2\) and the four sides have area \(4xh\)
   • \(S=x^2+4x\cdot\frac{4000}{x^2}=x^2+\frac{16000}{x}\)

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

   • Graph \(y=x^2+\frac{16000}{x}\) for x > 0
   • From the calculator: minimum at (20, 1200)
   • S = 1200 cm², when x = 20

A cinema sells 400 tickets for a film at €10 each. For every 50 cent rise in the price, 15 fewer tickets are sold. Each ticket costs the cinema €2. Let x be the number of 50 cent rises.

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

   • Price: \(10+0.5x\) euros
   • Number sold: \(400-15x\)

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

   • Profit per item: \(10+0.5x-2=8+0.5x\)
   • \(P=(8+0.5x)(400-15x)\)

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

   • Graph \(y=(8+0.5x)(400-15x)\). From the calculator: maximum at x = 5.33
   • x must be a whole number. 5: P = 3412.50; 6: P = 3410.00
   • x = 5 gives the larger profit. Price \(=10+0.5(5)=12.50\) euros; profit €3412.50

A swimming pool is a rectangle of length L metres with a semicircle of diameter w metres at each end. The distance all the way round the edge of the pool is 160 m.

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

   • \(2L+\pi w=160\)
   • \(L=\frac{160-\pi w}{2}\)

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

   • \(A=Lw=\frac{w(160-\pi w)}{2}\)

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

   • Graph \(y=\frac{x(160-\pi x)}{2}\) for \(0<x<160/\pi\)
   • From the calculator: maximum at (25.5, 1020)
   • A = 1020 m², when w = 25.5 m

A closed cylindrical pencil case has radius r cm and height h cm. Its total surface area is 900 cm².

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

   • \(2\pi r^2+2\pi rh=900\), so \(h=\frac{900-2\pi r^2}{2\pi r}\)
   • \(V=\pi r^2h=450r-\pi r^3\)

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

   • Graph \(y=450x-\pi x^3\) for x > 0
   • From the calculator: maximum at (6.91, 2070)
   • V = 2070 cm³, when r = 6.91 cm

A rectangular patio measuring x metres by y metres has an area of 450 m². It is edged with stone. The edging along the two sides of length x costs £10 per metre, and along the two sides of length y costs £4 per metre.

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

   • \(xy=450\), so \(y=\frac{450}{x}\)
   • \(C=20x+8y=20x+\frac{3600}{x}\)

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

   • Graph \(y=20x+\frac{3600}{x}\) for x > 0
   • From the calculator: minimum at (13.4, 537)
   • Least cost £536.66; x = 13.4 m and y = 33.5 m