Inequalities and linear programming  (8.48.6)
Represent the graphs of linear inequalities in two unknowns on a plane

Graph

Examples

Solutions
Solve systems of linear inequalities in two unknowns

Theory

ExamplesFind the maximum and minimum values of subject to the constraints \(\left\{ \begin{array}{l}3x + 11y  66 \le 0\\x + y  6 \ge 0\\5x  3y  46 \le 0\end{array} \right.\).

SolutionsThe 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

Graph

Theory

ExamplesA shop sells two different jewellery sets A and B. The relevant information of the two jewellery sets of is as follows.
Necklaces Watches Rings Selling Price ($/k) Set A 3 1 2 25 Set B 1 1 4 20
(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.

Solutionsa) 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 \(\begin{array}{l} = \$ 5(5x + 4y)\\ = \$ 5{(5 \times 25 + 4 \times 75)^{}}\\ = {\underline{\underline {\$ 2{\rm{ }}125}} ^{}}\end{array}\)