### Inequalities and linear programming - (8.4-8.6)

#### Solve systems of linear inequalities in two unknowns

• Find the maximum and minimum values of subject to the constraints $$\left\{ \begin{array}{l}3x + 11y - 66 \le 0\\x + y - 6 \ge 0\\5x - 3y - 46 \le 0\end{array} \right.$$.
• The feasible solutions of all constraints are represented graphically as below.

Let $$f(x{\rm{ }},{\rm{ }}y) = x + 2y$$.
Draw the straight line $$f(x{\rm{ }},{\rm{ }}y) = 0$$.
From the graph, $$f(x{\rm{ }},{\rm{ }}y)$$ attains its maximum / minimum values at the points $$(8{\rm{ }},{\rm{ }} - {\rm{ }}2)$$ and $$(11{\rm{ }},{\rm{ }}3)$$.
At the point $$(8{\rm{ }},{\rm{ }} - {\rm{ }}2)$$, $$f(8{\rm{ }},{\rm{ }} - {\rm{ }}2) = 8 + 2( - {\rm{ }}2) = 4$$.
At the point $$(11{\rm{ }},{\rm{ }}3)$$, $$f(11{\rm{ }},{\rm{ }}3) = 11 + 2(3) = 17$$.

Maximum value$$= \underline{\underline {17}}$$
Minimum value$$= \underline{\underline 4}$$

Alternative method:
Consider the vertices $$(0{\rm{ }},{\rm{ }}6)$$, $$(8{\rm{ }},{\rm{ }} - {\rm{ }}2)$$ and $$(11{\rm{ }},{\rm{ }}3)$$ of the feasible region.
At the point $$(0{\rm{ }},{\rm{ }}6)$$, $$f(0{\rm{ }},{\rm{ }}6) = 0 + 2(6) = 12$$.
At the point $$({\rm{8 }},{\rm{ }} - {\rm{ }}2)$$, $$f(8{\rm{ }},{\rm{ }} - {\rm{ }}2) = 8 + 2( - {\rm{ }}2) = 4$$.
At the point $$(11{\rm{ }},{\rm{ }}3)$$, $$f(11{\rm{ }},{\rm{ }}3) = 11 + 2(3) = 17$$.

Maximum value$$= \underline{\underline {{\rm{17}}}}$$
Minimum value$$= \underline{\underline {\rm{4}}}$$

#### Solve linear programming problems

• A shop sells two different jewellery sets A and B. The relevant information of the two jewellery sets of is as follows.
NecklacesWatchesRingsSelling Price (\$/k)
Set A31225
Set B11420
Mr. Chan wants to buy at least 150 necklaces, 100 watches and 240 rings.

(a) Write down all the constraints about the number of sets A and sets B to be bought.
(b) How many sets of each type should Mr. Chan buy to minimize the cost? Find the minimum cost.
• a) Suppose x sets A and y sets B are to be bought,the constraints are:
$$\left\{ \begin{array}{l}3x + y \ge 150\\x + y \ge {100^{}}\\2x + 4y \ge {240^{}}\\x{\rm{ \text{ and } }}y{\rm{ \text{ are non - negative integer} }}{{\rm{s}}^{}}\end{array} \right.$$,

which are equivalent to $$\left\{ \begin{array}{l}3x + y \ge 150\\x + y \ge {100^{}}\\x + 2y \ge {120^{}}\\x{\rm{ \text{ and } }}y{\rm{ \text{ are non - negative integer} }}{{\rm{s}}^{}}\end{array} \right.$$.

b) Cost $$\begin{array}{c}\ C(x{\rm{ }},{\rm{ }}y) = \ (25x + 20y)\\ = \ 5{(5x + 4y)^{}}\end{array}$$
Represent the feasible solutions graphically and draw the straight line $$5(5x + 4y) = 0$$, i.e. the straight line $$5x + 4y = 0$$.

The ordered pairs representing all points with integral coordinates in the shaded region represent the feasible solutions.

From the graph, the cost is the minimum when $$x = 25$$ and $$y = 75$$.
Mr. Chan should buy 25 sets A and 75 sets B.
Minimum cost $$\begin{array}{l} = \ 5(5x + 4y)\\ = \ 5{(5 \times 25 + 4 \times 75)^{}}\\ = {\underline{\underline {\ 2{\rm{ }}125}} ^{}}\end{array}$$