The process of finding the maximum or minimum value of a functions is called optimisation.
For the quadratic function $y=ax^2+bx+c$, we have already seen that the vertex has $x$-coordinate $-\dfrac{b}{2a}$.

$a>0$: the minimum value of $y$ occurs at $x=-\dfrac{b}{2a}$ $a<0$: the maximum value of $y$ occurs at $x=-\dfrac{b}{2a}$ There are many cases for which we need to identify the maximum or minimum value for a situation. The solution is often referred to as the optimum solution and the process is called optimisation.
There are several ways of finding optimum solutions including:

• graphing the function accurately and search for the maximum or minimum from the drawings
• using $\displaystyle x=-\dfrac{b}{2a}$ for finding its vertex, in case that the situation is relevant to a parabola
• applying differential calculus to locate the turning points Note: Not all of turning points is either maximum or minimum!
The maximum or minimum value highly depends on the domain. It is essential to examine the values of the function at the endpoints of the domain under consideration for global maxima and minima. Take a look at the following graph for your consideration.
However the local maximum value occurs at $x=b$ and the local minimum value occurs at $x=c$, they are neither a maximum nor a minimum in the given domain $a \le x \le d$. As you can see from the graph, $x=a$ gives the minimum value and $x=d$ gives the minimum value. ## Second Derivative Test

Local maximum: $\cap$ shape, $f^{\prime}(x) = 0$ and $f^{\prime \prime}(x) \lt 0$
Local minimum: $\cup$ shape, $f^{\prime}(x) = 0$ and $f^{\prime \prime}(x) \gt 0$

## Optimisation Processes

### Step 1

Draw a precise and clear diagram of the situation.

### Step 2

Construct an expression with the variable to be optimised as the problem. It should be written in terms of one convenient variable, say $x$. It should be written down what domain restrictions there are on $x$.

### Step 3

Find the first derivative to find out the turning points which could be either maximum or minimum value.

### Step 4

For a restricted domain, check whether the maximum value or minimum value is in the given domain. Check whether there are any other $x$-values could produce either global maximum or minimum value.

### Example 1

Find the maximum or minimum value of $y=x^2+6x-1$ and the corresponding value of $x$.

Since $a=1 \gt 0$, the shape of the graph is concave down and it has the minimum value. The minimum value occurs at:
\begin{align} \displaystyle x &= -\dfrac{b}{2a} \\ &= -\dfrac{6}{2 \times 1} \\ &= -3 \\ y &= (-3)^2 + 2 \times (-3) -1 \\ &= 2 \\ \end{align}
So the minimum value of $y$ is $2$, occurring at $x=-3$.

### Example 2

Find the maximum or minimum value of $y=-2x^2+8x+1$ and the corresponding value of $x$.

Since $a=-2 \lt 0$, the shape of the graph is concave up and it has the maximum value. The maximum value occurs at:
\begin{align} \displaystyle x &= -\dfrac{b}{2a} \\ &= -\dfrac{8}{2 \times (-2)} \\ &= 2 \\ y &= -2 \times 2^2 + 8 \times 2 + 1 \\ &= 9 \\ \end{align}
So the minimum value of $y$ is $9$, occurring at $x=2$.

### Example 3

The profit in selling $x$ computers per day, is given $P=-3x^2 + 120x – 400$ dollars. Find the maximum profit per day.

Since $a=-3 \lt 0$, the shape of the graph is concave up and it has the maximum value.
The maximum value occurs at:
\begin{align} \displaystyle x &= -\dfrac{b}{2a} \\ &= -\dfrac{120}{2 \times (-3)} \\ &= 20 \\ y &= -3 \times 20^2 + 120 \times 20 – 400 \\ &= 800 \\ \end{align}
So the maximum profit is $800$ dollars, occurring when $20$ computers are sold per day. 