Optimisation


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

Step 5

Write your answer in a sentence, making sure the answer is appropriate to the question.

Example 1

Find the maximum and minimum values of $y=x^2$, for $1 \le x \le 2$.


From the graph, its maximum value is $4$, when $x=2$, and its minimum value is $1$, when $x=1$.

Example 2

Find the maximum and minimum values of $y=x^2$, for $-2 \le x \le -1$.


From the graph, its maximum value is $4$, when $x=-2$, and its minimum value is $1$, when $x=-1$.

Example 3

Find the maximum and minimum values of $y=x^2$, for $-1 \le x \le 2$.


From the graph, its maximum value is $4$, when $x=2$, and its minimum value is $0$, when $x=0$.

Example 4

Find the maximum and minimum values of $y=x^2$, for $-2 \le x \le 1$.


From the graph, its maximum value is $4$, when $x=-2$, and its minimum value is $0$, when $x=0$.

Example 5

Consider the function $3x^4+4x^3-12x^2-3$ for $-3 \le x \le 2$.

(a)   Find the x values of the stationary points.

\( \begin{align} \displaystyle
f'(x) &= 12x^3+12x^2-24x \\
12x^3+12x^2-24x &= 0 \\
x^3+x^2-2x &= 0 \\
x(x^2+x-2)x &= 0 \\
x(x+2)(x-1) &=1 \\
\therefore x &= 0 \text{ or } x = x=-2 \text{ or } x=1 \\
\end{align} \)

(b)   Find the values of $y$ of the stationary points and endpoints.

For stationary points;
\( \begin{align} \displaystyle
f(-2) &= 3 \times (-2)^4 + 4 \times (-2)^3 – 12 \times (-2)^2 – 3 = -35 \\
f(0) &= 3 \times 0^4 + 4 \times 0^3 – 12 \times 0^2 – 3 = -3 \\
f(1) &= 3 \times 1^4 + 4 \times 1^3 – 12 \times 1^2 – 3 = -8 \\
\end{align} \)
For endpoints;
\( \begin{align} \displaystyle
f(-3) &= 3 \times (-3)^4 + 4 \times (-3)^3 – 12 \times (-3)^2 – 3 = 24 \\
f(2) &= 3 \times 2^4 + 4 \times 2^3 – 12 \times 2^2 – 3 = 29 \\
\end{align} \)

(c)   Find the maximum value.

$24$ at $x=-3$

(d)   Find the minimum value.

$-35$ at $x=-2$





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