LP PRACTICE QUESTION.docx

LP PRACTICE QUESTION.docx - A farmer can plant up to 8...

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A farmer can plant up to 8 acres of land with wheat and barley. He can earn $5,000 for every acre he plants with wheat and $3,000 for every acre he plants with barley. His use of a necessary pesticide is limited by federal regulations to 10 gallons for his entire 8 acres. Wheat requires 2 gallons of pesticide for every acre planted and barley requires just 1 gallon per acre. What is the maximum profit he can make? SOLUTION TO PROBLEM NUMBER 1 let x = the number of acres of wheat let y = the number of acres of barley. since the farmer earns $5,000 for each acre of wheat and $3,000 for each acre of barley, then the total profit the farmer can earn is 5000*x + 3000*y. let p = total profit that can be earned. your equation for profit becomes: p = 5000x + 3000y that's your objective function. it's what you want to maximize the constraints are: number of acres has to be greater than or equal to 0. number of acres has to be less than or equal to 8. amount of pesticide has to be less than or equal to 10. your constraint equations are: x >= 0 y >= 0 x + y <= 8 2x + y <= 10 to graph these equations, solve for y in those equations that have y in them and then graph the equality portion of those equations. x >= 0 y >= 0 y <= 8-x y <= 10 - 2x x = 0 is a vertical line that is the same line as the y-axis. y = 0 is a horizontal line that is the same line as the x-axis. the area of the graph that satisfies all the constraints is the region of feasibility.
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the maximum or minimum solutions to the problem will be at the intersection points of the lines that bound the region of feasibility. the graph of your equations looks like this: the region of feasibility is the shaded area of the graph. you can see from this graph that the region of feasibility is bounded by the following (x,y) coordinate points: (0,0) (0,8) (2,6) (5,0) the point (0,0) is the intersection of the line x-axis with the y-axis. the point (0,8) is the intersection of the line y = 8 - x with the y-axis. the point (5,0) is the intersection of the line y = 10 - 2x with the x-axis. the point (2,6) is the intersection of the line y = 8 - x with the line y = 10 - 2x. the point (2,6) was solved for in the following manner: equations of the intersecting lines are: y = 8 - x
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y = 10 - 2x subtract the first equation from the second equation and you get: 0 = 2 - x add x to both sides of this equation and you get: x = 2 substitute 2 for x in either equation to get y = 6. that makes the intersection point (x,y) = (2,6). the objective equation is: p = 5000x + 3000y profit will be maximum at the intersection points of the region of feasibility on the graph. the profit equation is evaluated at each of these points as shown in the following table. intersection point of (x,y) p (0,0) $0 (0,8) $24,000 (2,6) $28,000 ***** (5,0) $25,000 the maximum profit occurs when the farmer plants 2 acres of wheat and 6 acres of barley. number of acres of wheat is 2 and number of acres of barley is 6 for a total of 8 acres which is the maximum number of acres available for planting.
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