GENERAL MATHS – UNIT TWO

GENERAL MATHS – UNIT TWO
Linear Programming GENERAL MATHS – UNIT TWO

Review - Sketching Linear Graphs
The Gradient-Intercept Method The linear equation needs to be arranged in the form 𝒚=𝒎𝒙+𝒄 * Plot the y-intercept on the graph (0,𝑐) * Use ′𝑚′ (the gradient) to determine the next point to plot 𝑚= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 * Join the two points together

eg1. Use the gradient-intercept method to draw the graph of 𝒚= 𝟑 𝟐 𝒙+𝟏
The Gradient-Intercept Method The linear equation needs to be arranged in the form 𝒚=𝒎𝒙+𝒄 * Plot the y-intercept on the graph (0,𝑐) * Use ′𝑚′ (the gradient) to determine the next point to plot 𝑚= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 * Join the two points together eg1. Use the gradient-intercept method to draw the graph of 𝒚= 𝟑 𝟐 𝒙+𝟏 The y-intercept is 1, which gives point (0, 1) To sketch, we need one more point. We can find this from gradient, 𝑚= 3 2 = 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 This tells us, for every 3 points up (risen), we move across 2 points (run). (2, 4) (0, 1)

eg2. Use the gradient-intercept method to draw the graph of 𝒚=𝒙−𝟒
The Gradient-Intercept Method The linear equation needs to be arranged in the form 𝒚=𝒎𝒙+𝒄 * Plot the y-intercept on the graph (0,𝑐) * Use ′𝑚′ (the gradient) to determine the next point to plot 𝑚= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 * Join the two points together eg2. Use the gradient-intercept method to draw the graph of 𝒚=𝒙−𝟒 The y-intercept is -4, which gives point (0, −4) To sketch, we need one more point. We can find this from gradient, 𝑚=1= 1 1 = 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 This tells us, for every 1 point up (risen), we move across 1 point (run). (1, -3) (0, -4)

The y-intercept is -1, which gives point (0, −1)
The Gradient-Intercept Method The linear equation needs to be arranged in the form 𝒚=𝒎𝒙+𝒄 * Plot the y-intercept on the graph (0,𝑐) * Use ′𝑚′ (the gradient) to determine the next point to plot 𝑚= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 * Join the two points together eg3: Use the gradient-intercept method to draw the graph of 𝒚=𝟑𝒙−𝟏 The y-intercept is -1, which gives point (0, −1) To sketch, we need one more point. We can find this from gradient, 𝑚=3= 3 1 = 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 This tells us, for every 3 points up (risen), we move across 1 point (run). (1, 2) (0, -1)

Rearranging an equation from 𝑎𝑥+𝑏𝑦=𝑐 into 𝑦=𝑚𝑥+𝑐 form Rearrange 4𝑥+2𝑦=12 to 𝑦=𝑚𝑥+𝑐 form: We want to make y the subject: 2𝑦=−4𝑥+12 𝑦= −4𝑥+12 2 𝑦=−2𝑥+6

Rearranging an equation from 𝑎𝑥+𝑏𝑦=𝑐 into 𝑦=𝑚𝑥+𝑐 form Rearrange −6𝑥+3𝑦=15 to 𝑦=𝑚𝑥+𝑐 form: We want to make y the subject: 3𝑦=6𝑥+15 𝑦= 6𝑥+15 3 𝑦=2𝑥+5

The 𝑥−𝑦 Intercept Method Can be used for linear equations in either form 𝑦=𝑚𝑥+𝑐 or 𝑎𝑥+𝑏𝑦+𝑐=0 * Find 𝑦-intercept: Solve 𝑦 when 𝑥 =0 by substituting 𝑥=0 into the linear equation, this gives (0, 𝑦) * Find 𝑥-intercept: Solve 𝑥 when 𝑦 =0 by substituting 𝑦=0 into the linear equation, this gives 𝑥, 0 * Plot these two points on the graph and rule them together to form a line.

eg1. Sketch the line given by the equation 2𝑥+3𝑦=6 Let 𝑥=0: 2 0 +3𝑦=6
The 𝑥−𝑦 Intercept Method Can be used for linear equations in either form y = mx + c or ax + by = c * Find 𝑦-intercept: Solve 𝑦 when 𝑥 =0 by substituting 𝑥=0 into the linear equation, this gives (0, 𝑦) * Find 𝑥-intercept: Solve 𝑥 when 𝑦 =0 by substituting 𝑦=0 into the linear equation, this gives 𝑥, 0 * Plot these two points on the graph and rule them together to form a line. eg1. Sketch the line given by the equation 2𝑥+3𝑦=6 Let 𝑥=0: 2 0 +3𝑦=6 3𝑦=6 𝑦=2 (0, 2) Let 𝑦=0: 2𝑥+3(0)=6 2𝑥=6 𝑥=3 (3, 0) (0, 2) (3, 0)

eg2. Sketch the line given by the equation −4𝑥+2𝑦=8 Let 𝑥=0:
The 𝑥−𝑦 Intercept Method Can be used for linear equations in either form y = mx + c or ax + by = c * Find 𝑦-intercept: Solve 𝑦 when 𝑥 =0 by substituting 𝑥=0 into the linear equation, this gives (0, 𝑦) * Find 𝑥-intercept: Solve 𝑥 when 𝑦 =0 by substituting 𝑦=0 into the linear equation, this gives 𝑥, 0 * Plot these two points on the graph and rule them together to form a line. eg2. Sketch the line given by the equation −4𝑥+2𝑦=8 Let 𝑥=0: −4 0 +2𝑦=8 2𝑦=8 𝑦= (0, 4) Let 𝑦=0: −4𝑥+2(0)=8 −4𝑥=8 𝑥=−2 (−2, 0) (0, 4) (-2, 0)

eg3. Sketch the line given by the equation 𝑦=3𝑥+6 Let 𝑥=0: 𝑦=3𝑥+6
The 𝑥−𝑦 Intercept Method Can be used for linear equations in either form y = mx + c or ax + by = c * Find 𝑦-intercept: Solve 𝑦 when 𝑥 =0 by substituting 𝑥=0 into the linear equation, this gives (0, 𝑦) * Find 𝑥-intercept: Solve 𝑥 when 𝑦 =0 by substituting 𝑦=0 into the linear equation, this gives 𝑥, 0 * Plot these two points on the graph and rule them together to form a line. eg3. Sketch the line given by the equation 𝑦=3𝑥+6 Let 𝑥=0: 𝑦=3𝑥+6 𝑦=3 0 +6 𝑦= (0, 6) Let 𝑦=0: 0=3𝑥+6 −6=3𝑥 𝑥= −6 3 =− (−2, 0) (0, 6) (-2, 0)

eg. Sketch the line given by the equation 𝑦=4
In this topic you will also come across more basic graphs, involving only an 𝑥 or 𝑦 variable eg. Sketch the line given by the equation 𝑦=4 (0, 4)

eg2. Sketch the line given by the equation 𝑥=−6
In this topic you will also come across more basic graphs, involving only an 𝑥 or 𝑦 variable eg2. Sketch the line given by the equation 𝑥=−6 (-6, 0)

eg3. Sketch the line given by the equation 𝑦=0
In this topic you will also come across more basic graphs, involving only an 𝑥 or 𝑦 variable eg3. Sketch the line given by the equation 𝑦=0 (0, 0)

Now Try Worksheet One Page One

Using the calculator to sketch graphs 𝒙=𝟑 Highlight and drag into the graph box

Using the calculator to sketch graphs 𝒚=𝟐 Highlight and drag into the graph box

Using the calculator to sketch graphs 𝒚=𝟑𝒙+𝟔 To find the co-ordinates of the y-intercept: Analysis  G-Solve  y-intercept To find the co-ordinates of the x-intercept Analysis  G-Solve  root Highlight and drag into the graph box

Using the calculator to sketch graphs 𝟏𝟎𝒙+𝟐𝒚=𝟔 To find the co-ordinates of the y-intercept: Analysis  G-Solve  y-intercept To find the co-ordinates of the x-intercept Analysis  G-Solve  root Highlight and drag into the graph box

Using the calculator to sketch 2 graphs on one plot 𝟏𝟎𝒙+𝟐𝒚=𝟔 To find the co-ordinates of the Point of Intersection: Analysis  G-Solve  Intersect Highlight and drag each equation into the graph box 𝒚=−𝒙+𝟑

Now: Then: Check your answers to the worksheet using your calculator
Do the other side of the page Now: Then:

Understanding Inequalities
What do the following symbols represent? < > ≤ ≥ 𝑥< 𝑥 𝑖𝑠 𝑥> 𝑥 𝑖𝑠 𝑥≤ 𝑥 𝑖𝑠 𝑥≥− 𝑥 𝑖𝑠 −3 less than greater than less than or equal to greater than or equal to less than greater than less than or equal to greater than or equal to

How can we represent these values on a number line? 𝑥>3 𝑥<4 1<𝑥<7

What about on an x-y axis? Shade the region which isn’t included in the range. 𝑥> 𝑥≤ −2<𝑥<4 Check your answer by choosing a test point in the unshaded area and sub it into the equation.

What about on an x-y axis? Shade the region which isn’t included in the range. 𝑦> y≤ −6<𝑦<0 Check your answer by choosing a test point in the unshaded area and sub it into the equation.

What about on an x-y axis? Shade the region which isn’t included in the range. 𝑦> x≤− <𝑥<8 Check your answer by choosing a test point in the unshaded area and sub it into the equation.

Rearranging Inequations
We may be presented with inequations in one variable which require some rearranging first. −3𝑥+2≥8 −3𝑥≥6 −𝑥≥2 𝑥≤−2 2𝑥−4≥6 2𝑥≥10 𝑥≥5 𝑦+5<8 𝑦<3

Now Try Exercise 11.2 Questions 1, 2, 3, 4, 7

Plotting Inequalities with BOTH 𝑥 and 𝑦
eg1. Draw the graph of the inequation 𝒚<𝟐𝒙+𝟓 Plot the graph of 𝑦=2𝑥+5 Decide which side is defined by the equation by choosing a test point and checking if it satisfies the inequality equation 0<2 0 +5 0<0+5 0<5 So we can leave this side of the graph and shade the other side of the line Try (0, 0): Is this true? YES

eg2. Draw the graph of the inequation 𝒚−𝟒𝒙<𝟐 Plot the graph of 𝑦−4𝑥=2 Decide which side is defined by the equation by choosing a test point and checking if it satisfies the inequality equation 0−4(0)<2 0<2 So we can leave this side of the graph and shade the other side of the line Try (0, 0): Is this true? YES

eg3. Draw the graph of the inequation 𝒚+𝟑𝒙>𝟔 Plot the graph of 𝑦+3𝑥=6 Decide which side is defined by the equation by choosing a test point and checking if it satisfies the inequality equation 0+3 0 >6 0>6 So we can shade the side of the graph that includes the point ( 0, 0) Try (0, 0): Is this true? NO

eg4. Draw the graph of the inequation 𝟐𝒚+𝟒𝒙≥𝟔 Plot the graph of 2𝑦+4𝑥=6 Decide which side is defined by the equation by choosing a test point and checking if it satisfies the inequality equation ≥6 0≥6 So we can shade the side of the graph that includes the point ( 0, 0) Try (0, 0): Is this true? NO

eg5. Draw the graph of the inequation 𝒚<𝟐𝒙 Plot the graph of 𝑦=2𝑥 Decide which side is defined by the equation by choosing a test point and checking if it satisfies the inequality equation – AS THIS GRAPH GOES THROUGH THE ORIGIN, WE CANNOT USE (0,0) AS THE TEST POINT. 0<2 1 0<2 So we can leave this side of the graph and shade the other side of the line Try (1, 0): Is this true? YES

Finding the inequality, when given the equation & graph
Sometimes we are given the linear equation and given the graph, but have to fill in the inequality sign. eg1. Given the equation below, complete the inequality given the following graph… 𝒚 ⎕ 𝟐𝒙+𝟏 Substitute a test point into the equation (0, 0): 𝒚 ⎕ 𝟐𝒙+𝟏 𝟎 ⎕ 𝟐 𝟎 +𝟏 𝟎 ⎕ 𝟎+𝟏 𝟎 ⎕ 𝟏 Look at the graph - Is the test point in the R.R? If YES, then the last line in the equation above needs to be TRUE If NO, then the last line in the equation above needs to be FALSE This point is NOT in the R.R, so make the statement FALSE 𝟎 > 𝟏 Now substitute the inequality into the equation. 𝒚 > 𝟐𝒙+𝟏

Finding the inequality, when given the equation & graph
Sometimes we are given the linear equation and given the graph, but have to fill in the inequality sign. eg2. Given the equation below, complete the inequality given the following graph… 𝒚 ⎕ 𝟒𝒙+𝟐 Substitute a test point into the equation (0, 0): 𝒚 ⎕ 𝟒𝒙+𝟐 𝟎 ⎕ 𝟒 𝟎 +𝟐 𝟎 ⎕ 𝟎+𝟐 𝟎 ⎕ 𝟐 Look at the graph - Is the test point in the R.R? If YES, then the last line in the equation above needs to be TRUE If NO, then the last line in the equation above needs to be FALSE This point IS in the R.R, so make the statement TRUE 𝟎 < 𝟐 Now substitute the inequality into the equation. 𝒚 < 𝟒𝒙+𝟐

Now Try Exercise 11.2 Questions 5, 6, 8, 9, 10, 11, 13, 16

Finding an inequation, given the graph
eg1. Find the inequation that represents the graph Choose 2 points on the graph Find the gradient: 𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 Sub your gradient into ‘m’ in 𝑦=𝑚𝑥+𝑐 Sub the co-ordinates of any point into the 𝑦 and 𝑥 and rearrange to solve ‘𝑐’ 5. Sub the m and the c value into 𝑦=𝑚𝑥+𝑐 Or read ‘c’ off the graph Easier to choose intercepts: Points (-2, 0) and (0, 4) Sub points into 𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 4−0 0−−2 = =2 𝑦=2𝑥+𝑐 𝑆𝑢𝑏 𝑖𝑛 0,4 : =2 0 +𝑐 𝑐=4 5. 𝑦=2𝑥+4

Finding an inequation, given the graph
eg1 cont. Find the inequation that represents the graph 𝒚=𝟐𝒙+𝟒 Now the equals sign needs to be replaced with an inequality. Choose test point (0,0) – As this point is shaded, we need to choose an inequality that produces a FALSE…. 0=2 0 +4 0=4 0> 𝐹𝐴𝐿𝑆𝐸 −𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑤ℎ𝑎𝑡 𝑤𝑒 𝑤𝑎𝑛𝑡….. 𝑠𝑜 𝑦>2𝑥+4

Now Try Exercise 11.2 Questions 5, 6, 8, 9, 10, 11, 13, 16, 17b, 18a

Graphing two inequations on one plot
We plot 2 inequations on the same axis & find the region which satisfies both inequations. We will again shade the region not defined by the inequation, leaving the required area unshaded. Required region eg1. 𝒙≥𝟏 𝒚≥𝟑 Plot each of these on the axis Shade the area not defined by the inequation The remaining region satisfies both inequations

We will again shade the region not defined by the inequation, leaving the required area unshaded. eg2. 𝒙≤𝟎 𝒚>𝟓 Required region Plot each of these on the axis Shade the area not defined by the inequation The remaining region satisfies both inequations

We will again shade the region not defined by the inequation, leaving the required area unshaded. eg3. 𝒙<𝟎 𝒚>𝟐𝒙+𝟏 Required region Plot each of these on the axis Shade the area not defined by the inequation The remaining region satisfies both inequations Check your answer using a test point

𝒚<𝟐𝒙+𝟏 𝒚+𝒙≥𝟒 eg4. Sketch the pair of simultaneous inequations.
Label the point of intersection (POI) on your graph 𝒚<𝟐𝒙+𝟏 𝒚+𝒙≥𝟒 Required region Plot each of these on the axis Shade the area not defined by the inequation The remaining region satisfies both inequations Check your answer using a test point Highlight the Point of Intersection on the graph

𝒙−𝒚>𝟑 𝒚+𝒙≤𝟐 eg5. Sketch the pair of simultaneous inequations.
Label the point of intersection (POI) on your graph 𝒚+𝒙≤𝟐 𝒙−𝒚>𝟑 Required region Plot each of these on the axis Shade the area not defined by the inequation The remaining region satisfies both inequations Check your answer using a test point Highlight the Point of Intersection on the graph

Now Try Exercise 11.3 Questions 1, 2, 3, 4, 11, 12, 14

Graphing systems of inequations on one plot
We will again shade the region not defined by the inequation, leaving the required area unshaded. eg1. Find the required region defined by the system of inequations 𝒙≥𝟎 𝒚≥𝟏 𝒚<𝟖 𝒙≤𝟔 Required region Plot each of these on the axis Shade the area not defined by the inequation The remaining region satisfies all inequations

eg2. Find the required region defined by the system of inequations 𝒙≥𝟑 𝒙≤𝟕 𝒚≥𝟑 𝒚<𝟔 Required region Plot each of these on the axis Shade the area not defined by the inequation The remaining region satisfies all inequations

eg3. Find the required region defined by the system of inequations 𝒙≥𝟏 𝐲≤𝟐 𝒚<𝟑𝒙−𝟏 →𝑡𝑒𝑠𝑡 𝑝𝑜𝑖𝑛𝑡 0,0 : 0<−1 FALSE– So shade region which includes the point (0,0) Required region

eg4. Find the required region defined by the system of inequations 𝒙≥𝟏 𝒚≤−𝟐𝒙+𝟓 →𝑡𝑒𝑠𝑡 𝑝𝑜𝑖𝑛𝑡 0,0 : 0<5 TRUE– So leave region which includes the point (0,0) 𝒚<𝟐𝒙−𝟗 →𝑡𝑒𝑠𝑡 𝑝𝑜𝑖𝑛𝑡 0,0 : 0<−9 FALSE– So shade region which Required region

Now Try Exercise 11.3 Questions 1, 2, 3, 4, 11, 12, 14

Linear Programming Applications: Solving worded problems
These problems involve converting worded problems into a series of linear inequations. Using the corner points of the feasible region, we can solve real problems involving finding a maximum or minimum number – ie. Finding a maximum profit, or a minimum Cost. Objective function is the Profit or Cost equation that we are trying to maximise or minimise. Constraints are the linear inequations we create and sketch to find the feasible region. Constraints may include things like the amount of material we have to make something or the amount of time it takes to do something

eg1. A local factory produces runners and walking shoes. It is able to produce a minimum of 400 pairs of runners and 350 pairs of walking shoes. The factory is capable of producing a maximum of 900 pairs of shoes altogether. The profit on a pair of runners is $12.50 and on a pair of walking shoes, $10. What combination of shoes should the factory make to maximise profits? 1. Define to variables to solve your question: 2. Define the Objective Function: 3. Define the Constraints: 𝑥=𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑖𝑟𝑠 𝑜𝑓 𝑟𝑢𝑛𝑛𝑒𝑟𝑠 𝑦=𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑖𝑟𝑠 𝑜𝑓 𝑤𝑎𝑙𝑘𝑖𝑛𝑔 𝑠ℎ𝑜𝑒𝑠 𝑃=12.5𝑥+10𝑦 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑟𝑢𝑛𝑛𝑒𝑟𝑠: 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑤𝑎𝑙𝑘𝑖𝑛𝑔 𝑠ℎ𝑜𝑒𝑠: 𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑥𝑖𝑚𝑢𝑚: 𝑥≥400 𝑦≥350 𝑥+𝑦≤900

1. Define to variables to solve your question:
𝑥=𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑖𝑟𝑠 𝑜𝑓 𝑟𝑢𝑛𝑛𝑒𝑟𝑠 𝑦=𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑖𝑟𝑠 𝑜𝑓 𝑤𝑎𝑙𝑘𝑖𝑛𝑔 𝑠ℎ𝑜𝑒𝑠 1. Define to variables to solve your question: 2. Define the Objective Function: 3. Define the Constraints: 𝑃=12.5𝑥+10𝑦 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑟𝑢𝑛𝑛𝑒𝑟𝑠: ≤𝑥 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑤𝑎𝑙𝑘𝑖𝑛𝑔 𝑠ℎ𝑜𝑒𝑠: ≤𝑦 𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑥𝑖𝑚𝑢𝑚: ≥𝑥+𝑦 4. Plot the constraints and find the feasible region: 5. Find the corner points of the feasible region and label them: 𝐴: 400, 𝐵: 400, 𝐶: 550,350 6. Use the corner points to solve the objective function: 𝐴: 400, 𝑃= =8500 𝐵: 400, 𝑃= =10000 𝐶: 550, 𝑃= =10375 7. Answer the question given in the initial problem: We want to maximise profit, so if 550 pairs of runners and 350 pairs of walkers are produced and sold, the company makes a profit of $10375

eg2. A farmer has a plot of 35 hectare plot of land to plant oats and wheat. Oats require 3 hours of labour to produce per hectare. Wheat require 4 hours of labour to produce per hectare. A total of 120 hours of labour is available. Profit on oats is $200 per hectare and wheat is $240 per hectare. What should he plant to maximise profits? 1. Define to variables to solve your question: 2. Define the Objective Function: 3. Define the Constraints: 𝑥=ℎ𝑒𝑐𝑡𝑎𝑟𝑒 𝑜𝑓 𝑜𝑎𝑡𝑠 𝑦=ℎ𝑒𝑐𝑡𝑎𝑟𝑒 𝑜𝑓 𝑤ℎ𝑒𝑎𝑡 𝑃=200𝑥+240𝑦 35≥𝑥+𝑦 120≥3𝑥+4𝑦 𝑥≥0 𝑦≥0 𝑆𝑝𝑎𝑐𝑒 𝐴𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒: 𝐿𝑎𝑏𝑜𝑢𝑟:

2. Define the Objective Function: 3. Define the Constraints: 𝑥=ℎ𝑒𝑐𝑡𝑎𝑟𝑒 𝑜𝑓 𝑜𝑎𝑡𝑠 𝑦=ℎ𝑒𝑐𝑡𝑎𝑟𝑒 𝑜𝑓 𝑤ℎ𝑒𝑎𝑡 𝑃=200𝑥+240𝑦 𝑆𝑝𝑎𝑐𝑒 𝐴𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒: 35≥𝑥+𝑦 120≥3𝑥+4𝑦 𝑥≥0 𝑦≥0 𝐿𝑎𝑏𝑜𝑢𝑟: 4. Plot the constraints and find the feasible region: 5. Find the corner points of the feasible region and label them: 𝐴: 0, 𝐵: 0, 𝐶: 20, D:(35,0) 6. Use the corner points to solve the objective function: 𝐴: 0, 𝑃= =0 𝐵: 0, 𝑃= =0+7200=7200 𝐶: 20, 𝑃= = =7600 𝐷: 35, 𝑃= =7000 7. Answer the question given in the initial problem: We want to maximise profit, so if 20ha of oats and 15ha of wheat is produced, the farmer will achieve a maximum profit of $7600

Worksheet Exercise Now try then

Maximising and minimising linear functions
Using linear programming, we can maximise or minimise linear functions subject to constraints given by a set of linear inequations. Today we will do this using the corner point method, using the following steps: Sketch the feasible region Determine the coordinates of the corner points of the feasible region Substitute the coordinates of the corner points into the given objective function (the linear equation to be maximised or minimised) 4. Choose which set of coordinates produces the maximum or minimum value, as stated in the question.

𝑧=𝑥+2𝑦 eg1. Maximise 𝑧=𝑥+2𝑦 subject to the system of inequations
Sketch the feasible region Determine the coordinates of the corner points of the feasible region Sub. the corner points into the objective function (the linear equation to be maximised or minimised) Choose which set of coordinates produces the maximum or minimum value, as stated in the question eg1. Maximise 𝑧=𝑥+2𝑦 subject to the system of inequations 𝒙≥𝟎 𝒚≥𝟏 𝒚<𝟖 𝒙≤𝟔 Required region Corner Points (0, 1) (0, 8) (6, 1) (6, 8) 𝑧=𝑥+2𝑦 𝑧=0+(2×1) =2 𝑧=0+(2×8) =16 𝑧=6+(2×1) =8 𝑧=6+(2×8) =22 𝑍 𝑚𝑎𝑥 =22

𝑧=2𝑥−2𝑦 eg2. Minimise 𝑧=2𝑥−2𝑦 subject to the system of inequations
Sketch the feasible region Determine the coordinates of the corner points of the feasible region Sub. the corner points into the objective function (the linear equation to be maximised or minimised) Choose which set of coordinates produces the maximum or minimum value, as stated in the question eg2. Minimise 𝑧=2𝑥−2𝑦 subject to the system of inequations 𝒙≥𝟎 𝒚≥𝟑 𝒚<𝟏𝟎 𝒙≤𝟒 Required region Corner Points (0, 3) (0, 10) (4, 3) (4, 10) 𝑧=2𝑥−2𝑦 𝑧= 2×0 −(2×3) =−6 𝑧= 2×0 −(2×10) =−20 𝑧= 2×4 −(2×3) =8−6=2 𝑧= 2×4 −(2×10) =8−20=−12 𝑍 𝑚𝑖𝑛 =−20

𝑧=3𝑥+2𝑦 eg3. Maximise 𝑧=3𝑥+2𝑦 subject to the system of inequations
Sketch the feasible region Determine the coordinates of the corner points of the feasible region Sub. the corner points into the objective function (the linear equation to be maximised or minimised) Choose which set of coordinates produces the maximum or minimum value, as stated in the question eg3. Maximise 𝑧=3𝑥+2𝑦 subject to the system of inequations 𝒙≥𝟎 𝒚≥𝟎 𝟐𝒚+𝟐𝒙<𝟖 Required region Corner Points (0, 0) (0, 4) (4, 0) 𝑧=3𝑥+2𝑦 𝑧= 3×0 +(2×0) =0 𝑧= 3×0 +(2×4) =8 𝑧= 3×4 +(2×0) =12 𝑍 𝑚𝑎𝑥 =12

𝑧=4𝑥−𝑦 eg4. Minimise 𝑧=4𝑥−𝑦 subject to the system of inequations
Sketch the feasible region Determine the coordinates of the corner points of the feasible region Sub. the corner points into the objective function (the linear equation to be maximised or minimised) Choose which set of coordinates produces the maximum or minimum value, as stated in the question eg4. Minimise 𝑧=4𝑥−𝑦 subject to the system of inequations 𝒚≥−𝟔 𝒚≤𝟑𝒙−𝟔 𝟐𝒚+𝟒𝒙<𝟖 Required region Corner Points (2, 0) (0, -6) (6, -6) 𝑧=4𝑥−𝑦 𝑧= 4×2 −0 =8 𝑧= 4×0 −−6 =6 𝑧= 4×6 −−6 =24+6=30 𝑍 𝑚𝑖𝑛 =6

Maximum/Minimum Problems
Now Try Worksheet Maximum/Minimum Problems Then: Exercise 11.4: Q 8, 1, 2

Worksheet Solutions

We are asked to Maximise – the largest solution from above is 10.
eg1. For the following system of inequations: Maximise z = x + y subject to 𝑥 ≥ 0, 𝑦 ≥ 0, 𝑥 ≤ 4, 𝑦 ≤ 6 𝐴: 0, 0 → 𝑧=𝑥+𝑦 = 0+0 =0 𝐵: 0, 6 → 𝑧=𝑥+𝑦=0+6=6 𝐶: 4, 6 →𝑧=𝑥+𝑦=4+6=10 𝐷: 4,0 →𝑧=𝑥+𝑦=4+0=4 We are asked to Maximise – the largest solution from above is 10. So 𝑍 𝑚𝑎𝑥 =10

We are asked to Maximise – the largest solution from above is 15
eg2. For the following system of inequations: Maximise z = 2x + 3y subject to 𝑥 ≥ 0, 𝑦 ≥ 0, 𝑦≤2𝑥+3, 2𝑥+𝑦≤6 y ≤ 2x+3, 2x+y ≤ 6 𝐴: 0, 0 → 𝑧=2𝑥+3𝑦 = 0+0 =0 𝐵: 0, 3 → 𝑧=2𝑥+3𝑦=0+(3×3)=9 𝐶: 0.75, 4.5 →𝑧=2𝑥+3𝑦= 2×.75 +(3×4.5)=15 𝐷: 3,0 →𝑧=2𝑥+3𝑦=(2×3)+0=6 We are asked to Maximise – the largest solution from above is 15 So 𝑍 𝑚𝑎𝑥 =15

We are asked to Minimise – the smallest solution from above is - 4
eg3. For the following system of inequations: Minimise z = 2x - y subject to 𝑥 ≥ 0, 𝑦 ≥ 0, 𝑦 ≤ 4, 𝑥 ≤ 8 𝐴: 0, 0 → 𝑧=2𝑥−𝑦 = 0−0 =0 𝐵: 0, 4 → 𝑧=2𝑥−𝑦=0−4=−4 𝐶: 8, 4 →𝑧=2𝑥−𝑦=16−4=12 𝐷: 8,0 →𝑧=2𝑥−𝑦=16+0=16 We are asked to Minimise – the smallest solution from above is - 4 So 𝑍 𝑚𝑖𝑛 =−4

eg4. For the following system of inequations: Maximise z = x + 3y subject to
𝑥 ≥ 0, 𝑦 ≥ 0, 𝑥+𝑦≤8, 2𝑦−2𝑥≤4 𝐴: 0, 0 → 𝑧=𝑥+3𝑦 = 0+0 =0 𝐵: 0, 2 → 𝑧=𝑥+3𝑦=0+(3×2)=6 𝐶: 3, 5 →𝑧=𝑥+3𝑦= 3×3 + 3×5 =9+15=24 𝐷: 8,0 →𝑧=𝑥+3𝑦=(1×8)+0=8 We are asked to Maximise – the largest solution from above is 24 So 𝑍 𝑚𝑎𝑥 =24

eg1. A local factory produces runners and walking shoes. It is able to produce a minimum of 400 pairs of runners and 350 pairs of walking shoes. The factory is capable of producing a maximum of 900 pairs of shoes altogether. The profit on a pair of runners is $12.50 and on a pair of walking shoes, $10. What combination of shoes should the factory make to maximise profits? 1. Define to variables to solve your question: 2. Define the Objective Function: 3. Define the Constraints: 𝑥=𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑖𝑟𝑠 𝑜𝑓 𝑟𝑢𝑛𝑛𝑒𝑟𝑠 𝑦=𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑖𝑟𝑠 𝑜𝑓 𝑤𝑎𝑙𝑘𝑖𝑛𝑔 𝑠ℎ𝑜𝑒𝑠 𝑃=12.5𝑥+10𝑦 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑟𝑢𝑛𝑛𝑒𝑟𝑠: 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑤𝑎𝑙𝑘𝑖𝑛𝑔 𝑠ℎ𝑜𝑒𝑠: 𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑥𝑖𝑚𝑢𝑚: 𝑥≥400 𝑦≥350 𝑥+𝑦≤900

𝑥=𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑖𝑟𝑠 𝑜𝑓 𝑟𝑢𝑛𝑛𝑒𝑟𝑠 𝑦=𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑖𝑟𝑠 𝑜𝑓 𝑤𝑎𝑙𝑘𝑖𝑛𝑔 𝑠ℎ𝑜𝑒𝑠 1. Define to variables to solve your question: 2. Define the Objective Function: 3. Define the Constraints: 𝑃=12.5𝑥+10𝑦 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑟𝑢𝑛𝑛𝑒𝑟𝑠: ≤𝑥 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑤𝑎𝑙𝑘𝑖𝑛𝑔 𝑠ℎ𝑜𝑒𝑠: ≤𝑦 𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑥𝑖𝑚𝑢𝑚: ≥𝑥+𝑦 4. Plot the constraints and find the feasible region: 5. Find the corner points of the feasible region and label them: 𝐴: 400, 𝐵: 400, 𝐶: 550,350 6. Use the corner points to solve the objective function: 𝐴: 400, 𝑃= =8500 𝐵: 400, 𝑃= =10000 𝐶: 550, 𝑃= =10375 7. Answer the question given in the initial problem: We want to maximise profit, so if 550 pairs of runners and 350 pairs of walkers are produced and sold, the company makes a profit of $10375

eg2. A farmer has a plot of 35 hectare plot of land to plant oats and wheat. Oats require 3 hours of labour to produce per hectare. Wheat require 4 hours of labour to produce per hectare. A total of 120 hours of labour is available. Profit on oats is $200 per hectare and wheat is $240 per hectare. What should he plant to maximise profits? 1. Define to variables to solve your question: 2. Define the Objective Function: 3. Define the Constraints: 𝑥=ℎ𝑒𝑐𝑡𝑎𝑟𝑒 𝑜𝑓 𝑜𝑎𝑡𝑠 𝑦=ℎ𝑒𝑐𝑡𝑎𝑟𝑒 𝑜𝑓 𝑤ℎ𝑒𝑎𝑡 𝑃=200𝑥+240𝑦 35≥𝑥+𝑦 120≥3𝑥+4𝑦 𝑥≥0 𝑦≥0 𝑆𝑝𝑎𝑐𝑒 𝐴𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒: 𝐿𝑎𝑏𝑜𝑢𝑟:

2. Define the Objective Function: 3. Define the Constraints: 𝑥=ℎ𝑒𝑐𝑡𝑎𝑟𝑒 𝑜𝑓 𝑜𝑎𝑡𝑠 𝑦=ℎ𝑒𝑐𝑡𝑎𝑟𝑒 𝑜𝑓 𝑤ℎ𝑒𝑎𝑡 𝑃=200𝑥+240𝑦 𝑆𝑝𝑎𝑐𝑒 𝐴𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒: 35≥𝑥+𝑦 120≥3𝑥+4𝑦 𝑥≥0 𝑦≥0 𝐿𝑎𝑏𝑜𝑢𝑟: 4. Plot the constraints and find the feasible region: 5. Find the corner points of the feasible region and label them: 𝐴: 0, 𝐵: 0, 𝐶: 20, D:(35,0) 6. Use the corner points to solve the objective function: 𝐴: 0, 𝑃= =0 𝐵: 0, 𝑃= =0+7200=7200 𝐶: 20, 𝑃= = =7600 𝐷: 35, 𝑃= =7000 7. Answer the question given in the initial problem: We want to maximise profit, so if 20ha of oats and 15ha of wheat is produced, the farmer will achieve a maximum profit of $7600

Now Try Exercise 11.4 Q 5, 6, 7, 12, 13, 18

GENERAL MATHS – UNIT TWO

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