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Optimization of a Two Variable Function

Optimization of a Two Variable Function. Optimizing a function of x and y consists in finding the vertex of a polygon of constraints which , depending on the situation, maximizes or minimizes the function .

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Optimization of a Two Variable Function

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  1. Optimization of a Two Variable Function

  2. Optimizing a function of x and y consists in finding the vertex of a polygon of constraintswhich, depending on the situation, maximizes or minimizes the function. • In general, wetry to maximize a revenue function or minimize a costfunction • The optimizingfunctionis in the form: Z = ax + by + c, where a, b and c are numbers

  3. Steps to Solving an OptimizationFunction • Identify the variables • Translate the situation into a system of inequalities • Graph the polygon of constraints • Determine the vertices of the polygon of constraints • Establish the rule of the function to beoptimized (revenue or cost) • Evaluate the functionateach of the vertices • Deduce the vertex thatmaximizes or minimizes the function

  4. Example Ordinary gas sells for 4 ¢ less per litre than super gas at the corner gas station. A litre of super gas costs 83.9 ¢. This gas station sells at least twice as much ordinary gas as super gasoline. The total number of litres of gas sold daily never surpasses 8400, but the station never sells less than 1600 litres of super gas. The following system of inequalities describes the constraints of this situation. x 0 y 0 y 1600 x + y 8400 x 2y where:   x represents the number of litres of ordinary gas sold y represents the number of litres of super gas sold.  How many litres of each type of gas must be sold daily to maximize income?

  5. x 0 y 0 y 1600 x + y 8400 x 2y 8400 Vertices: (3200, 1600)(6800, 1600)(5600, 2800) 1600 400 400 8400 OptimizingFunctionZ = 0.799x + 0.839y

  6. Example: A steel mill converts steel into girders and rods. The mill can produce at most 100 units of steel a day. At least 20 girders and at least 60 rods are required daily. If the profit on a girder is $8 and the profit on a rod is $6, how many units of each type should the mill produce each day to maximize profits?

  7. Let x = number of girders Let y = number of rods

  8. x ≥ 20 y ≥ 60 x + y ≤ 100 100 10 10 100

  9. Vertices of the polygon: (20, 60) (20, 80) (40, 60) Optimizing Function Z = 8x + 6y

  10. 40 girders and 60 rods of steel should be produced for a maximum profit of $680.

  11. Example The manager of a warehouse store decided to offer his customers two models of a certain heat pump : the Thermex model and the Calorex model. The marketing department gathered the following information : • The annual demand for both models will be no more than 300 units. • The annual demand for both models will be at least 180 units. • A minimum of 100 Thermex models will likely be sold. • At most there are 60 more Calorex models than Thermex models in the warehouse. The company makes a profit of $525 on each Thermex model and $700 on each Calorex model. Let x : represent the number of Thermex models sold y : represent the number of Calorex models sold How many heat pumps of each model should be sold in order to maximize the profit?

  12. Constraints y  0 x  0 x + y  300 x + y 180 x  100 y x + 60

  13. Example The student council of a school is organizing a fashion show, which will be held in the school auditorium. The tickets will cost $4 for students at the school and $6 for the general public. The following polygon of constraints reflects the different constraints faced by the student council. x: number of tickets for students at the school y: number of tickets for the general public A runway will be built so that the models can walk down the middle of the auditorium. The runway will reduce the seating capacity. As a result, only a maximum of 240 tickets can be sold. By how much will the maximum possible revenue decrease once the runway is built?

  14. Maximum revenue before the runway is built Additional constraint

  15. Maximum revenue after the runway is built Decrease in maximum revenue: $1500  $1200 = $300

  16. Example Oliver works as a cashier and wrapper at a grocery store. There are different constraints that limit the number of hours he can work per week. These constraints are represented by the inequalities and the polygon of constraints given below. Each side of the polygon and its corresponding inequality are identified by the same number.y 4 x + y 16 y 4 + 2x x1 x: number of hours he can work per week as a cashier y: number of hours he can work perweek as a wrapper Oliver's boss suggests two ways of paying him: Option A: $13 per hour as a cashier and $8 an hour as a wrapper; Option B: $9 per hour, whether he works as a cashier or as a wrapper. Which option will allow Oliver to maximize his income?

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