iGCSE (2021 Edition)

11.11 Optimisation problems

Lesson

Differential calculus is often used in optimisation problems. That is, if we have a quantity that is varying according to a function (that can be differentiated) we can optimise that quantity by finding maximum and minimum values. We will look at applications involving area, volume, business, finance and motion.

Caution is required when dealing with optimisation problems and we need to undertake a sanity check on the solutions as they may not make sense for the application given. For example, a function may give negative values but if it was representing a volume or an area relationship those negative values would be nonsensical.

The general strategy for solving optimisation problems is:

1. Determine the quantity to be maximised or minimised
2. Determine a function for this quantity with respect to one variable
3. Differentiate the function to be optimised and find and classify maximum and minimum points accordingly
4. Complete a sanity check of your results

Worked example

Example 1

The manager of a computer shop has determined that the costs $C$C for storage and handling of new computers as a function of the number of computers $x$x is given by:

 $C$C $=$= $15x+\frac{24000}{x}+6000$15x+24000x​+6000

Determine how many computers the manager should order to minimise costs.

Think: We are asked to find the number of computers that minimises the storage and handling costs $C$C, so we will need to find a local minimum point for the function given.

Do: To find minimum turning points we need to find the first derivative of the function and determine the location of any stationary points.  We can then use the second derivative to classify the points.

Finding the first derivative of the cost function gives:

 $C'$C′ $=$= $15-\frac{24000}{x^2}$15−24000x2​

Solving $C'=0$C=0 to find the location of any stationary points:

 $15-\frac{24000}{x^2}$15−24000x2​ $=$= $0$0 $\frac{24000}{x^2}$24000x2​ $=$= $15$15 $x^2$x2 $=$= $\frac{24000}{15}$2400015​ $x^2$x2 $=$= $1600$1600 $x$x $=$= $\pm40$±40

At this point we can discount $x=-40$x=40 as we cannot order a negative number of computers.

Let's now find the second derivative to determine the nature of the stationary point at $x=40$x=40:

 $C''$C′′ $=$= $\frac{48000}{x^3}$48000x3​

At $x=40$x=40$C''$C will be positive and hence this point is a local minimum as the curve is concave up at this point. Therefore, ordering $40$40 computers will minimise storage and handling costs.

Practice questions

Question 1

A manufacturer of mens shirts determines that her profit $P$P varies with respect to the number of shirts produced according to $P=21x-0.2x^{\frac{3}{2}}-500$P=21x0.2x32500.

1. Find the derivative, $P'$P.

2. What is the $x$x-value of the stationary point?

3. Find the second derivative, $P''$P.

4. Which of the following is correct about the stationary point?

The stationary point is a point of inflection because the second derivative at this point is zero.

A

The stationary point is a minimum because the second derivative at this point is positive.

B

The stationary point is a maximum because the second derivative at this point is negative.

C

The stationary point is a point of inflection because the second derivative at this point is zero.

A

The stationary point is a minimum because the second derivative at this point is positive.

B

The stationary point is a maximum because the second derivative at this point is negative.

C
5. Therefore what number of shirts corresponds to a maximum profit?

Question 2

A pool is being emptied, and the volume of water $V$V litres left in the pool after $t$t minutes is given by the equation $V=1500\left(11-t\right)^3$V=1500(11t)3, for $0\le t\le11$0t11.

1. State the rate of change of the volume after $t$t minutes.

2. At what rate is the volume of water in the pool changing after $10$10 minutes?

3. Select the time $t$t at which the pool is emptying at the fastest rate.

$t=11$t=11 mins

A

$t=5$t=5 mins

B

$t=0$t=0 mins

C

$t=10$t=10 mins

D

$t=11$t=11 mins

A

$t=5$t=5 mins

B

$t=0$t=0 mins

C

$t=10$t=10 mins

D

Question 3

A rectangular cereal box with a square base and open top is to have a volume of $256$256 cm3.

1. If the side lengths of the base measure $x$x cm, and the height of the box measures $h$h cm, express $h$h in terms of $x$x.

2. Let the surface area of the open box be represented by $S$S.

Find an equation for $S$S in terms of $x$x only, simplifying where possible.

3. Now find the possible value(s) of $x$x that will minimise the amount of material required to make the box. Give your answer(s) to the nearest integer if necessary.

4. Complete the table to prove that when $x=8$x=8 cm, the least packaging material is used.

Give both values to the nearest integer.

 $x$x $\frac{dS}{dx}$dSdx​ $7$7 $8$8 $9$9 $\editable{}$ $0$0 $\editable{}$

Motion and optimisation

We can use the techniques of optimisation in the study of motion of a body to find maximum or minimum values for displacement, velocity, and acceleration of a body.

Velocity is the rate of change of displacement and acceleration is the rate of change of the velocity. So given the displacement as a function of time, we are able to determine functions for velocity and acceleration of the body using the following relationships:

Instantaneous velocity and acceleration

For a displacement function $x(t)$x(t) the instantaneous velocity $v(t)$v(t)  is the derivative of the displacement function with respect to time:

$v(t)=\frac{dx}{dt}$v(t)=dxdt

The instantaneous acceleration $a(t)$a(t)  is the derivative of the velocity function with respect to time:

$a(t)=\frac{dv}{dt}$a(t)=dvdt

Practice questions

Question 4

A particle $P$P starts from rest at point $O$O, and its velocity $t$t seconds after it starts moving is modeled by the equation $v=t^2\left(9-t\right)$v=t2(9t). After $9$9 seconds it comes to rest and stops moving.

1. Determine an equation for the acceleration $a$a of the particle after $t$t seconds.

Outcomes

0606C14.5E

Apply differentiation to practical maxima and minima problems.