Taking the derivative of the function lets us find critical points where the slope is \(0\) or undefined. These points tell us where the function might have a maximum or a minimum value. This approach is useful in many real-world situations, like optimising resources or finding the best design for a project. Learn how to use derivatives to find maxima and minima in functions.
Finding maxima and minima
The maximum or minimum values of a function occur where the derivative is \(0\). This is where the graph of the function has a horizontal tangent.
This means that if we find where a function has a derivative of \(0\), we can locate the maxima (highest point of a curve) and minima (lowest point of a curve).
Mathematically, we are solving for \(x\) when:
\[f'(x)=0\]
Second derivative test
As you might be able to tell from its name, the second derivative is the derivative of the derivative. For \(f(x)\), the derivative is \(f'(x)\) and the second derivative is \(f''(x)\).
Depending on the size of the second derivative, we can tell whether the \(x\) value we have calculated is an inflection point (where the curve changes from concave to convex or vice versa), a local maximum or local minimum.
For \(x\) where \(f'(x)=0\), if:
\[f''(x) \begin{cases} = 0,x\textrm{ is an inflection point}\\
<0,f(x)\textrm{ is a local maximum}\\
>0,f(x)\textrm{ is a local minimum}
\end{cases}\]
The second derivative for a function \(y\) in terms of \(x\) can also be written as \(\dfrac{d^{2}y}{dx^{2}}\).
Example 1 – finding maxima and minima
The distance \(s\textrm{ km}\) of a fishing boat from port at any time \(t\) hours, to the nearest kilometre, is given by the formula:
\[s=2+8t-2.5t^{2}\]
When is the boat furthest from port and what is its distance from the port at that time?
The furthest distance from the port is the \(s\) value when \(s'=0\). We need to solve for \(t\).
First, we find \(s'\).
\[s'=8-5t\]
Now, we can solve \(s'=0\) for \(t\).
\[\begin{align*} s' & = 0\\
8-5t & = 0\\
8 & = 5t\\
t & = \frac{8}{5}\\
& = 1.6\textrm{ h}
\end{align*}\]
We can check that this is a maximum distance using the second derivative test.
\[\begin{align*} s'' & = -5\\
& < 0
\end{align*}\]
\(t=1.6\) is therefore a maximum.
To find the distance from the port at that time, we substitute \(t=1.6\) into \(s\).
\[\begin{align*} s & = 2+8t-2.5t^{2}\\
& = 2+8(1.6)-2.5(1.6)^{2}\\
& = 8.4\textrm{ km}
\end{align*}\]
To summarise, the boat is furthest from port after \(1.6\textrm{ h}\), where it is \(8.4\textrm{ km}\) away.
Find the minimum value of the function \(f(x)=x^{2}-5x+6\).
Find the derivative \(f'(x)\) and solve for \(x\) if \(f'(x)=0\).
\[\begin{align*} f'(x) & = 2x-5\\
& = 0\\
2x & = 5\\
x & =\frac{5}{2}
\end{align*}\]
We can double check that it is a minimum.
\[\begin{align*} f''(x) & = 2\\
& > 0
\end{align*}\]
To find the minimum value of the function, we substitute \(x=\dfrac{5}{2}\) into \(f(x)\).
\[\begin{align*} f(x) & = x^{2}-5x+6\\
f\left(\frac{5}{2}\right) & = \left(\frac{5}{2}\right)^{2}-5\left(\frac{5}{2}\right)+6\\
& = \frac{25}{4}-\frac{25}{2}+6\\
& = \frac{25}{4}-\frac{50}{4}+\frac{24}{4}\\
& = -\frac{1}{4}
\end{align*}\]
Find the maximum product of two numbers that have a sum of \(10\).
Finding maxima and minima is not limited to graphs. Here, we can let the numbers be \(a\) and \(b\). The numbers have a sum of \(10\), so:
\[a+b=10\]
Let the product of the two numbers be \(P\), so:
\[P=a\times b\]
We need to get \(P\) in terms of \(a\) or \(b\) so that we are able to find derivative like \(\dfrac{dP}{da}\) or \(\dfrac{dP}{db}\). In this case, we will choose to write \(P\) as a function of \(b\), but we get the same result if we write it in terms of \(a\).
\[a=10-b\]
We can substitute this into the equation for \(P\).
\[\begin{align*} P & = (10-b)b\\
& = 10b-b^{2}
\end{align*}\]
We can now find the derivative of \(P\) with respect to \(b\).
\[\frac{dP}{db}=10-2b\]
For a maximum or minimum, \(\dfrac{dP}{db}=0\), and we can solve for \(b\).
\[\begin{align*} 10-2b & = 0\\
2b & = 10\\
b & = 5
\end{align*}\]
Again, we can check that it is a maximum.
\[\begin{align*} \dfrac{d^{2}P}{db^{2}} & = -2\\
& < 0
\end{align*}\]
We can substitute \(b=5\) into \(a+b=10\) to find \(a\).
\[\begin{align*} a+b & = 10\\
a+5 & = 10\\
a & = 5
\end{align*}\]
The question asks for the maximum product, so:
\[\begin{align*} P & = a\times b\\
& = 5\times5\\
& = 25
\end{align*}\]
Exercise – finding maxima and minima
Find the turning point of the parabola defined by \(y=f(x)=5x^{2}-30x+17\).
Find two positive numbers whose sum is \(18\) such that the sum of their squares is a minimum.
What is the maximum area that can be enclosed if a rectangle is created with a piece of wire \(48\textrm{ cm}\) long?
The annual profit \(P\) made on a garment is related to the number \(n\) that are produced by the formula \(P(n)=300n-7200-0.2n^{2}\). How many garments should be produced to maximise profit?
