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# PROGRAMMING - PowerPoint PPT Presentation

Linear programming techniques are used to solve a wide variety of problems, such as optimising airline scheduling and establishing telephone lines. PROGRAMMING. LINEAR.

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Linear programming techniques are used to solve a wide variety of problems, such as optimising airline scheduling and establishing telephone lines.

PROGRAMMING

LINEAR

• Next we will come up with inequalities that model the conditions or constraints for the situation.

• We'll then graph the constraints to come up with a region that satisfies all constraints.

• Find the vertices or corners of the region because these will be the possible values of x and y that maximize or minimize the objective function.

• We'll substitute these values in the objective function to obtain the maximum or minimum value depending on our situation.

• The x and y value that correspond to the maximum value (or minimum value) will be our solution.

A television manufacturer makes console and widescreen televisions. The profit per unit is \$71 for console televisions and \$127 for widescreen televisions. Equipment in the factory allows for the manufacturer to make at most 430 console televisions and 362 widescreen televisions per month. Also, the cost to the manufacturer per unit is \$493 for console televisions and \$633 for the widescreen televisions. Total cost per month may not exceed \$206,000. Determine the maximum profit and when it occurs.

Let's let c be the number of consoles built per month and w be the number of widescreens built per month.

Profit would be: 71c+127w

• First we will write an objective function. We desire to maximise the profit.

A television manufacturer makes console and widescreen televisions. The profit per unit is \$71 for console televisions and \$127 for widescreen televisions. Equipment in the factory allows for the manufacturer to make at most 430 console televisions and 362 widescreen televisions per month. Also, the cost to the manufacturer per unit is \$493 for console televisions and \$633 for the widescreen televisions. Total cost per month may not exceed \$206,000.

c = number of consoles w = number of widescreens

c 430

Constraint due to limit on number of consoles that can be built

Constraint due to limit on number of widescreens that can be built

w 362

493c + 633w 206000

Constraint due to limit on costs

c≥ 0 and w ≥ 0

• Next we will come up with inequalities that model the conditions or constraints for the situation.

c televisions. The profit per unit is \$71 for console televisions and \$127 for widescreen televisions. Equipment in the factory allows for the manufacturer to make at most 430 console televisions and 362 widescreen televisions per month. Also, the cost to the manufacturer per unit is \$493 for console televisions and \$633 for the widescreen televisions. Total cost per month may not exceed \$206,000.  430

c 430

w 362

493c + 633w 206000

w 362

493c + 633w 206000

Easiest to plot this line by finding c and w intercepts

493(0) + 633w = 206000

w 325

(0, 325)

493c + 633(0)  206000

(418,0)

c 418

So this is the area where we shaded for all three.

c≥ 0 and w ≥ 0

• We'll then graph the constraints to come up with a region that satisfies all constraints.

w televisions. The profit per unit is \$71 for console televisions and \$127 for widescreen televisions. Equipment in the factory allows for the manufacturer to make at most 430 console televisions and 362 widescreen televisions per month. Also, the cost to the manufacturer per unit is \$493 for console televisions and \$633 for the widescreen televisions. Total cost per month may not exceed \$206,000.  362

493c + 633w 206000

c 430

These vertices were easy because they were along the axes. If a vertice is not on an axis, the vertice is the intersection of two lines so solve the two equations together to find the point of intersection and hence the corner.

Vertice(0, 325)

Vertice(430, 0)

Vertice(0, 0)

• Find the vertices of the region because these will be the possible values of c and w that maximise or minimise the objective function.

This was our objective function. Our objective is to maximize profit.

P = 71c+127w

Vertice(430, 0)

Vertice(0, 325)

Vertice(0, 0)

P = 71(0)+127(0)

P = 71(430)+127(0)

P = 71(0)+127(325)

P = 0

P = 41,275

P = 30,530

Since this is the largest value, to maximize profit would be at the point (0, 325). This means do not make any console TV's and make 325 widescreens each month.

• The c and w value that correspond to the maximum value (or minimum value) will be our solution.

• We'll substitute these values in the objective function to obtain the maximum or minimum value depending on our situation.

Acknowledgement maximize profit.

I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint.

www.slcc.edu

Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.

Stephen Corcoran