Optimization is used to find the greatest/least value(s) a function can take. This can involve creating the expression first. Also find the rate of change by differentiating then substituting.

Part of

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.

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

Triangle AFE is shaded blue.

E lies on BC \(x\) units from B.

F lies on CD \(2x\) units from C.

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

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.

\[Are{a_{rectangle}} = \text{length} \times \text{breadth}\]

\[= 11 \times 7\]

\[= 77unit{s^2}\]

\[Are{a_{FCE}} = \frac{1}{2}bh\]

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

\[= x(7 - x)unit{s^2}\]

\[Are{a_{ABE}} = \frac{1}{2}bh\]

\[= \frac{1}{2} \times 11 \times x\]

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

\[Are{a_{DFE}} = \frac{1}{2}bh\]

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

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

Shaded area:

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

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

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

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 \(\frac{{dy}}{{dx}} = 0\).

\[2x - \frac{{11}}{2} = 0\]

\[4x - 11 = 0\]

\[4x = 11\]

\[x = \frac{{11}}{4}\]

\(\frac{{dy}}{{dx}} = 2(2) - \frac{{11}}{2} = - \frac{3}{2}\) (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 \(x = \frac{{11}}{4}\).