# Optimisation

To find the maximum or minimum values of a function, we would usually draw the graph in order to see the shape of the curve. Now, using our knowledge from differentiation, we can find these greatest and least values of a function without plotting the graph in a given interval.

## Example 1

A rectangle ABCD measures 7 units by 11 units, as shown.

E lies on BC $$x$$ units from B.

F lies on CD $$2x$$ units from C.

1. Show that the area, A units2, of the shaded region is given by $$A = {x^2} - \frac{{11}}{2}x + \frac{{77}}{2}$$
2. Find the value of $$x$$ for which the shaded area is least

### Solution

1. In order to find the shaded area, we would need to find the area of the rectangle then subtract the areas of the surrounding triangles.

$Are{a_{rectangle}} = \text{length} \times \text{breadth}$

$= 11 \times 7$

$= 77unit{s^2}$

$Are{a_{FCE}} = \frac{1}{2}bh$

$= \frac{1}{2} \times 2x \times (7 - x)$

$= x(7 - x)unit{s^2}$

$Are{a_{ABE}} = \frac{1}{2}bh$

$= \frac{1}{2} \times 11 \times x$

$= \frac{{11}}{2}x\,unit{s^2}$

$Are{a_{DFE}} = \frac{1}{2}bh$

$= \frac{1}{2} \times 7 \times (11 - 2x)$

$= \frac{7}{2}(11 - 2x)\,unit{s^2}$

$= 77 - (x(7 - x)) - \frac{{11}}{2}x - \left( {\frac{7}{2}(11 - 2x)} \right)$

$= 77 - 7x + {x^2} - \frac{{11}}{2}x - \frac{{77}}{2} + 7x$

$= {x^2} - \frac{{11}}{2}x + \frac{{77}}{2}\,unit{s^2}$

2. In order to find the least value of $$x$$, we need to find which value of $$x$$ gives us a minimum turning point. Therefore we need to differentiate and solve to find $$x$$, then find the nature in order to prove a minimum.

$A(x) = {x^2} - \frac{{11}}{2}x + \frac{{77}}{2}$

$A\textquotesingle(x)=2x-\frac{{11}}{2}$

Stationary points occur when $$\frac{{dy}}{{dx}} = 0$$.

$2x - \frac{{11}}{2} = 0$

$4x - 11 = 0$

$4x = 11$

$x = \frac{{11}}{4}$

### Nature

$$\frac{{dy}}{{dx}} = 2(2) - \frac{{11}}{2} = - \frac{3}{2}$$ (negative)

$$\frac{{dy}}{{dx}} = 2\left( {\frac{{11}}{4}} \right) - \frac{{11}}{2} = 0$$ (stationary)

$$\frac{{dy}}{{dx}} = 2(3) - \frac{{11}}{2} = \frac{1}{2}$$ (positive)

Therefore the surface area would be least when $$x = \frac{{11}}{4}$$.