Ronald Hui Tak Sun Secondary School HKDSE Mathematics Ronald Hui Tak Sun Secondary School

Missing Homework SHW2-A1 SHW2-C1 RE2 SHW2-R1 SHW2-P1 Sam L Sam L, Marco S RE2 Kelvin SHW2-R1 Tashi, Matthew, Kelvin, Sam L, Charles, Marco S, Pako, Yan Tin SHW2-P1 Tashi, Daniel, Matthew, Kelvin, Sam L, Ronald, Charles, Marco S, Marco W Missing Homework Ronald HUI 22 October 2015

Missing Homework SHW3-01 SHW3-A1 SHW3-B1 SHW3-C1 Daniel, Matthew, Ken, Kelvin, Sam L, Charles, Issac, Marco SHW3-A1 Daniel, Matthew, Ken, Sam L, Charles, Isaac, Yan Tin(RD) SHW3-B1 Tashi, Matthew, Charles, Isaac SHW3-C1 Tashi, Daniel, Matthew, Nathan, Treddick, Owen, Ken, Kelvin(RD), Sam L, Charles, Isaac, Marco S, Pako, Hanki, Marco W, Enoch Missing Homework Ronald HUI 22 October 2015

Missing Homework SHW3-D1 SHW3-E1 SHW3-R1 SHW3-P1 Tashi, Matthew, Kelvin(RD), Sam L, Isaac, Pako(RD), Aston(RD), Marco W, Enoch SHW3-E1 Tashi, Jason, Matthew, Nathan, Ken, Kelvin, Ronald, Charles, Isaac, Marco S, Hanki, Enoch SHW3-R1 Tashi, Daniel, Matthew, Nathan, Treddick(RD), Charles, Isaac, Marco S, Enoch SHW3-P1 Tashi, Jason, Matthew, Nathan, Ken, Kelvin, Ronald, Charles, Issac, Marco S, Hanki, Marco W, Enoch Missing Homework Ronald HUI 22 October 2015

Linear Programming

Linear Function in Two Variables A linear function in two variables x and y is a function in the form P = ax + by + c  The value of P depends on the values of x and y. where a, b and c are constants, a and b are not both zero. For example, P = x – 2y is a linear function in x and y.

If the values of x and y are restricted, then the value of P will also be restricted. Consider the linear function P = x + y. If x > 0 and y > 0, then P = x + y > 0.

Each restriction on the values of x and y is called a constraint. Consider the linear function P = x + y. If x > 0 and y > 0, ◄ Constraints then P = x + y > 0.

Constraints are usually expressed in the form of inequalities. Consider the linear function P = x + y. If x > 0 and y > 0, ◄ Constraints then P = x + y > 0.

An ordered pair (x, y) that satisfies all the given constraints is called a feasible solution.

Suppose a linear function P is subject to the constraints . ï î í ì ³ £ + 6 2 3 y x Consider two ordered pairs (0, 2) and (1, 1). For (0, 2), For (1, 1), 3x + 2y = 3(0)+ 2(2) = 4  6 3x + 2y = 3(1)+ 2(1) = 5  6 We can obtain all the feasible solutions by solving the set of inequalities graphically. x = 0  0 x = 1  0 y = 2  0 y = 1  0 ∵ (0, 2) and (1, 1) satisfy all the given constraints. ∴ They are parts of the feasible solutions.

Suppose a linear function P is subject to the constraints . ï î í ì ³ £ + 6 2 3 y x Graphical representation of : ï î í ì ³ £ + 6 2 3 y x

The feasible solutions are represented by the shaded region. The region that contains all the feasible solutions is called the feasible region. The feasible solutions are represented by the shaded region. Graphical representation of : ï î í ì ³ £ + 6 2 3 y x feasible region

Example: Find the minimum value of the function P = x + y subject to . Usually, we want to find the optimal value (the maximum value or the minimum value) of a linear function subject to some given constraints. Example: Find the minimum value of the function P = x + y subject to . ï î í ì ³ £ + 6 2 3 y x

Example: Find the minimum value of the function P = x + y subject to . We call the function an objective function. Example: Find the minimum value of the function P = x + y subject to . ï î í ì ³ £ + 6 2 3 y x Objective function

Example: Find the minimum value of the function P = x + y subject to . The feasible solution that optimizes (maximizes or minimizes) the value of an objective function is called the optimal solution of the function. Example: Find the minimum value of the function P = x + y subject to . ï î í ì ³ £ + 6 2 3 y x Objective function

Example: Find the minimum value of the function P = x + y subject to . ï î í ì ³ £ + 6 2 3 y x ∵ Among all the feasible solutions, (0, 0) minimizes the value of the objective function P = x + y. ∴ (0, 0) is an optimal solution for P. This type of problems belongs to a branch of Mathematics called linear programming. The corresponding optimal value (i.e. minimum value) is P = 0 + 0 = 0.

Consider the objective function P = x + y under the constraints . î í ì £  2 x y –2 Consider the objective function P = x + y under the constraints . feasible region of the given constraints x + y = 0 Consider the line x + y = 0 first.

At any point on the line x + y = 0, the value of x + y (i.e. P) is 0. î í ì £  2 x y –2 Consider the objective function P = x + y under the constraints . x + y = 0 At any point on the line x + y = 0, the value of x + y (i.e. P) is 0.

Similarly, at any point on the line x + y = 2, the value of P is 2. î í ì £  2 x y –2 Consider the objective function P = x + y under the constraints . x + y = 2 x + y = 0 Similarly, at any point on the line x + y = 2, the value of P is 2.

At any point on the line x + y = 4, the value of P is 4. î í ì £  2 x y –2 Consider the objective function P = x + y under the constraints . x + y = 2 x + y = 0 x + y = 4 At any point on the line x + y = 4, the value of P is 4.

At any point on the line x + y = 6, the value of P is 6. î í ì £  2 x y –2 Consider the objective function P = x + y under the constraints . ◄ They are parallel lines. x + y = 2 x + y = 0 x + y = 4 x + y = 6 At any point on the line x + y = 6, the value of P is 6.

Consider the objective function P = x + y under the constraints . î í ì £  2 x y –2 Consider the objective function P = x + y under the constraints . Value of P increases x + y = 2 x + y = 0 x + y = 4 x + y = 6 When the line x + y = 0 is translated to the right, the value of P increases.

Consider the following lines: î í ì £  2 x y –2 Consider the objective function P = x + y under the constraints . x + y = 0 x + y = –2 x + y = –4 ◄ They are parallel lines. x + y = –6 Consider the following lines: x + y = –2, x + y = –4 and x + y = –6.

Consider the objective function P = x + y under the constraints . î í ì £  2 x y –2 Consider the objective function P = x + y under the constraints . Value of P decreases x + y = 0 x + y = –2 x + y = –4 x + y = –6 When the line x + y = 0 is translated to the left, the value of P decreases.

To find the optimal values of an objective function that has a positive coefficient of x, such as P = x + y, the steps can be summarized as follows.

Step 1 Find the maximum and minimum values of the function P = x + y subject to the constraints . î í ì £  2 x y –2 Step 1 Draw and shade the feasible region.

Step 2 Draw the line x + y = 0 on the same coordinate plane. x + y = 0

Step 3 Translate the line x + y = 0 to the right within the feasible region to find the ordered pair (x, y) at which P attains its maximum. x + y = 4 x + y = 0 (2, 2) From the graph, P attains its maximum at (2, 2). ∴ Maximum value of P = 2 + 2 = 4

Step 4 Translate the line x + y = 0 to the left within the feasible region to find the ordered pair (x, y) at which P attains its minimum. x + y = 0 x + y = –4 (–2, –2) From the graph, P attains its minimum at (–2, –2). ∴ Minimum value of P = –2 + (–2) = –4

The optimal values of P can be found as follows. How if the objective function has a negative coefficient of x, like P = –x + y? The optimal values of P can be found as follows.

Find the maximum and minimum values of the function P = –x + y subject to the constraints . î í ì £  2 x y –2 –x + y = 0 ◄ Step 1: Draw and shade the feasible region. ◄ Step 2: Draw the line –x + y = 0 on the same coordinate plane.

Find the maximum and minimum values of the function P = –x + y subject to the constraints . î í ì £  2 x y –2 –x + y = 4 –x + y = 0 (–2, 2) ◄ Step 3: Translate the line –x + y = 0 to the left within the feasible region. From the graph, P attains its maximum at (–2, 2). ∴ Maximum value of P = –(–2) + 2 = 4

Find the maximum and minimum values of the function P = –x + y subject to the constraints . î í ì £  2 x y –2 –x + y = 0 –x + y = –4 ◄ Step 4: Translate the line –x + y = 0 to the right within the feasible region. (2, –2) From the graph, P attains its minimum at (2, –2). ∴ Minimum value of P = –2 + (–2) = –4

Example: Find the maximum value of the function P = x + y ï î í ì ³ £ 6 2 3 y x Step 1 subject to . (0, 3) Step 3 Step 2 From the graph, P attains its maximum at (0, 3). ∴ Maximum value of x + y = 0

Example: Find the maximum value of the function P = x + y ï î í ì ³ £ 6 2 3 y x subject to . In general, an objective function attains its optimal value at one of the vertices of the feasible region.

Example: Find the maximum value of the function P = x + y ï î í ì ³ £ 6 2 3 y x subject to . Therefore, we can test the values of P at all the vertices to find the optimal value.

Example: Find the maximum value of the function P = x + y ï î í ì ³ £ 6 2 3 y x subject to . The vertices of the feasible region are (0, 0), (0, 3) and (2, 0). At (0, 0), P = 0 + 0 = 0 At (0, 3), P = 0 + 3 = 3 At (2, 0), P = 2 + 0 = 2 ∴ The maximum value of P is 3.

Example: Find the maximum value of the function P = x + y ï î í ì ³ £ 6 2 3 y x subject to . Note that this method can only be used when all the vertices are feasible solutions.

Follow-up question (a) Draw and shade the region that satisfies the following constraints: ï î í ì - ³ £ 4 2 3 x y (b) Find the maximum and the minimum values of the function P = x + y – 1 subject to the constraints in (a).

(a) Graphically representation of : ï î í ì - ³ £ 4 2 3 x y (a) Graphically representation of :

From the graph, P attains its maximum at (3, 2). (b) (3, 2) ◄ Draw the line x + y – 1 = 0 on the same coordinate plane. x + y – 1 = 0 From the graph, P attains its maximum at (3, 2). ∴ Maximum value of

From the graph, P attains its minimum at (0, 0). (b) (0, 0) x + y – 1 = 0 From the graph, P attains its minimum at (0, 0). ∴ Minimum value of

Applications of Linear Programming

Delivery companies want to minimize their transportation costs. Linear programming is useful in different business sectors involving maximization of profits or minimization of costs under certain constraints. Example: Factory owners want to maximize their profits from fixed amounts of resources. Delivery companies want to minimize their transportation costs.

Let us illustrate how to solve real-life problems through the following example.

Jeff decides to spend at most 8 hours on working in a coffee shop and tutoring everyday. The time spent on tutoring should be at least 2 hours and at most 5 hours. If Jeff can earn $50 on each hour of working in the coffee shop and $100 on each hour of tutoring, find the number of hours he should spend on the two jobs each day to maximize his earning. What is the maximum amount can he earn each day? Step 1 Identify the unknown quantities and represent them by letters, say x and y. Let x and y be the numbers of hours Jeff spends on working in the coffee shop and tutoring respectively.

Jeff decides to spend at most 8 hours on working in a coffee shop and tutoring everyday. The time spent on tutoring should be at least 2 hours and at most 5 hours. If Jeff can earn $50 on each hour of working in the coffee shop and $100 on each hour of tutoring, find the number of hours he should spend on the two jobs each day to maximize his earning. What is the maximum amount can he earn each day? The constraints are: ï î í ì £ + 8 y x Step 2 Identify all constraints and express them as inequalities in terms of x and y. £ 5 2 y ³ y x ◄ The numbers of hours on working in the coffee shop and tutoring must be non-negative real numbers.

Jeff decides to spend at most 8 hours on working in a coffee shop and tutoring everyday. The time spent on tutoring should be at least 2 hours and at most 5 hours. Step 3 Draw and shade the region that satisfies all the constraints on a coordinate plane. If Jeff can earn $50 on each hour of working in the coffee shop and $100 on each hour of tutoring, find the number of hours he should spend on the two jobs each day to maximize his earning. What is the maximum amount can he earn each day? The constraints are: ï î í ì £ + 8 y x £ 5 2 y ³ y x

Express the objective function in terms of x and y. Jeff decides to spend at most 8 hours on working in a coffee shop and tutoring everyday. The time spent on tutoring should be at least 2 hours and at most 5 hours. If Jeff can earn $50 on each hour of working in the coffee shop and $100 on each hour of tutoring, find the number of hours he should spend on the two jobs each day to maximize his earning. What is the maximum amount can he earn each day? Step 4 Express the objective function in terms of x and y. Let $A be the amount he earns each day, then A = 50x + 100y.

Step 5 Find the maximum value of the objective function subject to the constraints. From the graph, A attains its maximum at (3, 5). (3, 5) ∴ Maximum value of A = 50(3) + 100(5) = 650 50x + 100y = 0 ∴ Jeff should spend 3 hours on working in the coffee shop and 5 hours on tutoring. The maximum amount he can earn is $650.

Follow-up question The following table shows the contents of phosphorus and nitrogen in 1 kg of fertilizers P and Q. Phosphorus Nitrogen Fertilizer P 3 units 2 units Fertilizer Q 4 units A farmer needs at least 40 units of phosphorus and 40 units of nitrogen for his crops. Fertilizer P costs $2/kg and fertilizer Q costs $3/kg. Suppose the farmer buys x kg of fertilizer P and y kg of fertilizer Q.

(a) Write down all the constraints on x and y. Draw and shade the region that satisfies the constraints in (a) on a coordinate plane. (c) Find the weights of fertilizer P and fertilizer Q that should be bought in order to minimize the cost. What is the minimum cost?

(a) The constraints are: Amount of Phosphorus Amount of Nitrogen x kg of Fertilizer P 3x units 2x units y kg of Fertilizer Q 2y units 4y units Minimum amount 40 units ï î í ì ³ + 40 4 2 3 y x (b)

(c) Let $C be the cost, then C = 2x + 3y. From the graph, C attains its minimum at (10, 5). ∴ Minimum value of C = 2(10) + 3(5) = 35 (10, 5) 2x + 3y = 0 ∴ The farmer should buy 10 kg of fertilizer P and 5 kg of fertilizer Q to attain a minimum cost of $35.