Algebra II: Strand 2. Linear Functions; Topic 4. Applications of Linear Programming; Task 2.4.1

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1 1 TASK 2.4.1: COTTON VS. CORN Solutions Suppose a farmer has 150 acres available for planting corn and cotton. The cotton seeds cost $3 per acre and the corn seeds cost $5 per acre. The total labor costs for cotton will be $15 per acre and the total labor costs for corn will be $8 per acre. The farmer expects the income from cotton to be $80 per acre and the income from corn to be $110 per acre. The farmer can spend no more than $540 on seeds and $1800 on labor. How many acres of corn and how many acres of cotton should the farmer plant in order to maximize his income? 1. Create a table to organize the information above. Cost of seeds/acre Cost of labor/acre Expected income/acre Cotton $3 $15 $80 Corn $5 $8 $110 Maximum allowed $540 $ What are the variables in this situation? Assign names to these variables. The number of acres of cotton seeds planted and the number of acres of corn seeds planted. We will use x and y to name these variables so that we can graph them easily. Let x = number of acres of cotton seeds planted. Let y = number of acres of corn seeds planted. 3. The farmer wants to maximize his income, I. Write an expression for I in terms of the variables you define in part 2. In linear programming, we call the function to be maximized or minimized the objective function. In this case, I is the objective function. I = 80x + 110y 4. Write a system of linear inequalities in x and y that describe the constraints on this situation. Connect each inequality with the problem situation. # 3x + 5y! 540 cost of seeds cannot exceed $540 % 15x + 8y! 1800 number of acres cannot exceed 1800 % $ x " 0 positive number of acres of cotton % y " 0 positive number of acres of corn % &% x + y! 150 total acres cannot exceed Rewrite each of the inequalities above in slope-intercept form (if possible) and graph the system. The feasible region represents all points that satisfy the system. Shade the feasible region on your graph and sketch below. Pick a point in your feasible region and show that it satisfies all of the constraints.

2 2 y (-3/5)x +108, y (-15/8)x + 225, y! 150 " x, x # 0, y # Pick some random points in the feasible region and have participants show how they meet the constraints of the problem. For each point you pick, ask participants to calculate the income earned. 6. What are the vertices of your feasible region? There are four vertices. One is the origin, one is the intersection of the two slanted lines and two are the intercepts of these lines. (0, 0), (0, 108), (120, 0), (91.76, 52.94) Do these points satisfy the constraints? Yes, because the inequalities are not strict inequalities, but include the boundaries. 7. Pick some points in the feasible region and fill in the table below with the income for the various points. Then, pick some points outside the feasible region and fill in the table provided. When comparing the two tables, what do you notice? Explain. Be sure that participants include all of the vertices of the region. They may choose

3 3 other points at random. We will discuss why the vertices are sufficient for maximizing the objective function in question 9. Points I = 80x + 110y (0, 0) I = 0 (0, 108) I = (120, 0) I = 9600 (91.76, 52.94) I = Participants should try points outside the region (for example, (150, 100) gives an income of $33,000). Picking values of x and y outside the feasible region may give a greater income, but the constraints are not met. 8. Refer to your first table (points in the feasible region) in question 7; use the values you obtained for income to find all points in the feasible region that could provide the specified income. Graph these equations on a graphing calculator. What do you notice about them? Explain. All of the lines have slope!8. That is, they are all parallel. For example, the income 11 found from planting 0 acres of cotton and 108 acres of corn, $9600, produces the function y = x. The income from planting acres of cotton and acres of corn, $ , produces the function y = x. Since we are maximizing our income, the optimal solution is the line that gives the highest y- intercept while still intersecting the feasible region. This is why we only really need to check the vertices of our region. If dynamic geometry software is available, show the Linear Programming model sketch. In it, participants see a feasible region with a possible income function. Drag the income line so that it is inside the feasible region and maximizes the income. What point of the feasible region lies on the line?

4 4 120 ( )!x f( x ) = g( x ) = -( )!x+225 Income function: y = -0.72x LINEAR PROGRAMMING MODEL Drag the Income function to maximize the income. Make sure that at least one point on the Income function is within the feasible region. y=f(x) 100 Income = C Income function 20 y=g(x) How many acres of cotton seed and how many acres of corn seed should the farmer plant in order to maximize his income? Will the farmer plant all 150 acres? Why or why not? He should plant 91.7 acres of cotton and 52.9 acres of corn. Why do we round down? If we round as we would mathematically, the solution would be 92 acres of cotton and 53 acres of corn. This violates one of our constraints since 3(92)+5(53)=541. The farmer cannot plant all 150 acres because this is outside the feasible region. After participants finish this activity, split them into groups to work on the Student Activity: Applications of Linear Programming.

5 5 Reflect and apply A trucker hauls citrus from Florida to New York City. Below is a chart of information giving the constraints and profit for the amount of fruit to be hauled. There is a company regulation that prohibits the driver from carrying more crates of grapefruit than crates of oranges. Volume in cubic ft Weight in lb Profit/crate Crates of oranges (x) 4 80 $2.50 Crates of $4.00 grapefruit (y) Maximum capacity of truck The trucker wants to maximize his profit. Below is his solution. $ x " 0 & y " 0 & % y # x & 4 x + 6y # 300 '& 80x +100y # P=.250x+4.00y 30 (30, 30) 20 (45, 20) 10 (0, 0) The maximum profit is $195 with 30 crates of oranges and 30 crates of grapefruit. A disease infects a large number of orange trees, significantly decreasing the amount of oranges available. So, the price of oranges begins to rise. The company takes off the regulation that a trucker cannot carry more grapefruit than oranges and the profit on oranges increases to $2.80 per crate. How does this affect the solution to the original problem? What does the profit on oranges need to be in order for a trucker to make a maximum profit by hauling only oranges? If the regulation is removed, the trucker should carry a full load of grapefruit to make the maximum profit. He can carry 50 crates and still meet his weight and volume restrictions. When the profit on oranges reaches $2.86 per crate, it becomes more beneficial for the trucker to carry a full load of oranges. Even though the profit is still less per crate, the trucker can carry more crates of (70, 0)

6 6 oranges and still meet his weight and volume restrictions. He can carry 70 crates of oranges. Notice that $2.86 is the solution to the equation 70x=(50)(4.00). Math notes In this task, participants reinforce the TEKS statements by looking at a graphical way of solving systems of equations or inequalities to solve a problem situation. Linear programming is a great example of mathematical modeling that is used in industry. Looking at constraints on solution sets is an important skill that will be useful in precalculus as students use functions to model situations. Teaching notes This task could be taken to the classroom as a class activity led by the teacher to introduce linear programming. Students would need a more detailed discussion of the system of linear inequalities in question 4. Sketching the graph of the feasible region would also take some guidance. If you wish to use the Inequal Application on the TI-83 plus, introduce this during question 5. Or, after participants have sketched these by hand, the facilitator can demonstrate the use of inequal to achieve the results. Students should explore more points in the feasible region to see if they satisfy each of the constraints. This can be done in question 7. Technology notes The application for the TI-83 plus can be used in place of the Geometer s Sketchpad. Inequal is preloaded on the Silver Edition calculators and can be downloaded free from the Texas Instruments Educational web site. Enter the inequalities as equations. Place the cursor over the equal sign. Press ALPHA (green key) ZOOM to change the equal sign to a less than or equal to sign.

7 7 Press ALPHA GRAPH to change the equal sign to a greater than or equal to sign. Arrow up to X= in the top left corner of the screen. Press ENTER. Enter x greater than or equal to zero. Now we need to set the window. Press WINDOW and set according to the screen shots. It is important to change the shade resolution to 6 since we are graphing four inequalities. Smaller resolutions make the screen too dark. Now press GRAPH. The inequalities will be graphed and shaded.

8 8 For a clearer view of the feasible region, press ALPHA, F1(Y=). Choose 1: Ineq Intersection. ALPHA ZOOM allows you to trace the points of intersection. You will need to use both the vertical and horizontal arrows to navigate to the points desired. Each time you hit a relevant point of intersection, press STO, ENTER. The x and y values will be stored in lists named LINEQX and LINEQY. To see the lists, press STAT, ENTER We can evaluate the objective function by using the empty list. Highlight the list and name it OBJ.

9 9 With the cursor over the list name OBJ, type in the objective function. Use quotes when typing in the formula for OBJ using the quotes indicates that you would like the graphing calculator to remember the formula otherwise the graphing calculator uses the formula to calculate values and only stores the values (i.e. it forgets the original function used to obtain the values). For x, use INEQX (2 nd LIST, arrow down to INEQX, ENTER). For y, use INEQY. We can now see which values of x and y maximize the objective function. When you are finished with the INEQUAL application, you must turn it off. Turning the calculator off does not disengage the application.

10 10 TASK 2.4.1: COTTON VS. CORN Suppose a farmer has 150 acres available for planting corn and cotton. The cotton seeds cost $3 per acre and the corn seeds cost $5 per acre. The total labor costs for cotton will be $15 per acre and the total labor costs for corn will be $8 per acre. The farmer expects the income from cotton to be $80 per acre and the income from corn to be $110 per acre. The farmer can spend no more than $540 on seeds and $1800 on labor. How many acres of corn and how many acres of cotton should the farmer plant in order to maximize his income? 1. Create a table to organize the information above. 2. What are the variables in this situation? Assign names to these variables. 3. The farmer wants to maximize his income, I. Write an expression for I in terms of the variables you define in part 2. In linear programming, we call the function to be maximized or minimized the objective function. In this case, I is the objective function. 4. Write a system of linear inequalities in x and y that describe the constraints on this situation. Connect each inequality with the problem situation.

11 11 5. Rewrite each of the inequalities above in slope-intercept form (if possible) and graph the system. The feasible region represents all points that satisfy the system. Shade the feasible region on your graph and sketch below. Pick a point in your feasible region and show that it satisfies all of the constraints. 6. What are the vertices of your feasible region? 7. Pick some points in the feasible region and fill in the table below with the income for the various points. Then, pick some points outside the feasible region and fill in the table provided. When comparing the two tables, what do you notice? Explain. Points in feasible region I =

12 12 Points outside feasible region I = 8. Refer to your first table (points in the feasible region) in question 7; use the values you obtained for income to find all points in the feasible region that could provide the specified income. Graph these equations on a graphing calculator. What do you notice about them? Explain. 9. How many acres of cotton seed and how many acres of corn seed should the farmer plant in order to maximize his income? Will the farmer plant all 150 acres? Why or why not?

13 13 Reflect and apply A trucker hauls citrus from Florida to New York City. Below is a chart of information giving the constraints and profit for the amount of fruit to be hauled. There is a company regulation that prohibits the driver from carrying more crates of grapefruit than crates of oranges. Crates of oranges (x) Crates of grapefruit (y) Maximum capacity of truck Volume in cubic ft Weight in lb Profit/crate 4 80 $ $ The trucker wants to maximize his profit. Below is his solution. $ x " 0 & y " 0 & 50 % y # x & 4 x + 6y # 300 P=.250x+4.00y '& 80x +100y # (30, 30) 20 (45, 20) 10 (0, 0) The maximum profit is $195 with 30 crates of oranges and 30 crates of grapefruit. (70, 0) A disease infects a large number of orange trees, significantly decreasing the amount of oranges available. So, the price of oranges begins to rise. The company takes off the regulation that a trucker cannot carry more grapefruit than oranges and the profit on oranges increases to $2.80 per crate. How does this affect the solution to the original problem? What does the profit on oranges need to be in order for a trucker to make a maximum profit by hauling only oranges?