Optimisation (Business)
Stationary Points
We are often interested in when the output of a function is at its lowest or highest possible value (maximum and minimum respectively).
A stationary point of a function is a point where the derivative of a function is equal to zero. Equivalently, it is a point where the slope of the graph of the function is zero (a straight line). A stationary point can be a minimum, maximum, or a point of inflection.
We can find the stationary points of a function $f(x)$ using the following method:
1) Find the derivative of the function with respect to $x$.
2) Set the derivative equal to zero \[\dfrac{\mathrm{d}f}{\mathrm{d}x}=0.\]
3) Solve the equation $\dfrac{\mathrm{d}f}{\mathrm{d}x}=0$ for $x$.This equation has one solution for each stationary point. Substituting each of these values (separately) into the function $f(x)$ will give the corresponding $y$ coordinate(s) of the stationary point(s).
4) Determine whether each stationary point is a maximum, a minimum or a point of inflection (see how to do this below).
Maximum or Minimum?
Using the Second Derivative
Once we have found the turning points of a function we then want to know whether this point is a maximum or a minimum or neither. One way to do this is to find the second derivative by differentiating the function twice. The second derivative is denoted by
\[\frac{ \mathrm{d}^2 f }{ \mathrm{d}x^2 } \text{ or } f''(x).\]
Suppose that we have a turning point at $x=a$. Then if:
- The second derivative is negative: $\frac{ \mathrm{d}^2 f }{ \mathrm{d}x^2 } \lt 0$ at $x=a$, we have a maximum.
- The second derivative is positive: $\frac{ \mathrm{d}^2 f }{ \mathrm{d}x^2 } \gt 0$ at $x=a$, we have a minimum.
Note: If the second derivative is zero: $\frac{ \mathrm{d}^2 f }{ \mathrm{d}x^2 }=0$ at $x=a$, the test is inconclusive: the point could be a maximum, a minimum or a point of inflection.
For more detailed information on derivatives and turning points, see Stationary Points.
Using the Sign of the First Derivative
Another method used to classify turning points is to look at the sign of the first derivative $\dfrac{\mathrm{d}}{\mathrm{d}x}(f(x))$ on either side of the turning point. If:
- The first derivative is positive on the left hand side of the turning point and negative on the right hand side, we have a maximum.
- The first derivative is negative on the left hand side of the turning point and positive on the right hand side, we have a minimum.
- In any other case, the derivative is neither a minimum nor a maximum.
Worked Example
Worked Example
Suppose that a company which sells calculators wishes to maximise profit. The company’s daily profit $P$ (in $£$), is given by $P=100q-0.1q^2$ where $q$ is the quantity of calculators sold per day. How many calculators should the company sell to maximise profit and what is the maximum daily profit?
Solution
To determine the quantity $q$ of calculators which maximises profit we must first find the derivative of the profit function $P(q)=100q-0.1q^2$ and set it equal to zero as follows:
\[\dfrac{\mathrm{d}P}{\mathrm{d}q}=100-0.2q\]
so we require $100-0.2q=0$.
We must now find the value of $q$ that makes this true. To do this we must first rearrange this equation so that only $q$ is on the left hand side:
\begin{align} 100-0.2q&=0\\ -0.2q&=-100\\ q&=\dfrac{100}{0.2}\\ &=500 \end{align}.
So the function $P(q)=100q-0.1q^2$ has a turning point when $q=500$, but is this a maximum? To answer this question we must find the second derivative of the function $P(q)=100q-0.1q^2$ and look at its sign (positive or negative) at $q=500$:
\[\dfrac{\mathrm{d}P}{\mathrm{d}q}=-0.2.\]
As the second derivative does not depend on $q$ (it is a constant), it will be equal to $-0.2$ (negative) for all values of $q$. As $-0.2$ is negative, we have found a maximum and so the company should sell $500$ calculators per day to maximise profit. Another way to see that $q=500$ is a maximum is to plot the function $P(q)=100q-0.1q^2$.
We can see from the graph that the function $P(q)=100q-0.1q^2$ takes the highest value when $q=500$ and that the slope changes from being positive when $q\lt500$ to negative when $q\gt500$. We now wish to find the maximum daily profit for the company. To do this we substitute the profit-maximising quantity of calculators sold, $q=500$, into the profit equation $P(q)=100q-0.1q^2$ as follows:
\begin{align} P(5)&=100×500-0.1×(500)^2\\ &=50,000-25,000\\ &=25,000\\ \end{align}
so the company’s maximum daily profit is $£25,000$.
Test Yourself
Test yourself: Numbas test on differentiation
Test yourself: Numbas test on differentiation, including the chain, product and quotient rules