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# Baseball Bat Company- Linear Programming

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## Jackson Rudoff

on 27 March 2014

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#### Transcript of Baseball Bat Company- Linear Programming

The Premier Pro Baseball Bat Company
Scenario
The City of Columbus has commissioned you, a former Major League Baseball player, to start a baseball bat company based in Columbus to compete with Louisville's chief export. They first give you \$10,000 to start off with production, but then request that neither of your two starting products cost more than \$7,500 dollars to manufacture.
You have two materials to work with, ash and maple. Ash bats are cheaper, only costing you \$20 to produce, but you will only sell for about \$40. Maple bats, however, will sell for about \$70, but will cost about \$45 to produce. As a side limitation, the city has asked that you never produce more maple than ash bats, in order to keep up with standard demand.
For solving this problem, let
x
represent the amount of ash bats and
y
represent the amount of maple bats. What combination of ash and maple bats would bring in the greatest profit?
Constraints
Objective
You need to find the maximum achievable profits from the given constraints. You know that ash bats will sell for \$40, while costing \$20 to produce, and maple will sell for \$70, while costing \$45 to produce. Ash bats are represented by x and maple bats will be represented by y. Therefore our objective function would be...

By Jackson Rudoff
Must keep cost under \$10,000
20x+45y<=10,000-------> y<=222.22-4/9x
Can't spend more than \$7,500 on one type of bat
20x<=7,500-------> x<=7,500
45y<=7,500-------> y<=7,500
Can never have more maple than ash
x<=y
Common Sense
x=>0
y=>0
Graph
f(x,y)=20x+25y
Variables
x=# of ash bats
y=# of maple bats
P= profits
Feasible Region
(375, 55.56)
(375, 0)
(153.85, 153.85)
(0,0)
(Graph is on a scale of 0-500 in intervals of 25)
Applying our Points
1- P(0, 0)= 20(0)+25(0)= \$0 (MIN)
2- P(154, 154)= 20(154)+25(154)= \$6,930
3- P(375, 56)= 20(375)+25(56)= \$8,900 (MAX)
4- P(375, 0)= 20(375)+25(0)= \$7,500
Before we can apply all of our intersection points we'll have to round a few up.

153.85----->154
55.56------>56
P(x, y)= 20x+25y
Solution
Making 375 ash bats and 56 maples bats would generate the greatest profit within the constraints given by the city.
(375, 56)= Maximum Value
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