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Question 251739: A company makes three products, A, B, and C. There are 500 pounds of raw material available. Each unit of product A requires 2 pounds of raw material, each unit of product B requires 2 pounds, and each unit of product C requires 3 pounds. The assembly line has 1,000 hours of operation available. Each unit of product A requires 4 hours, while each unit of products B and C requires 5 hours. The company realizes a profit of $500 for each unit of product A, $600 for each unit of product B, and $1,000 for each unit of product C. How many units of each of the three products should the company make to maximize profits?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! a = number of units of product A
b = number of units of product B
c = number of units of product C
raw material equation:
2a + 2b + 3c = 500 (equation 1)
hours of production equation:
4a + 5b + 5c = 1000 (equation 2)
profit equation:
500*a + 600*b + 1000*c = z (profit equation)
we have to solve equation 1 and 2 simultaneously.
we get:
2a + 2b + 3c = 500 (equation 1)
4a + 5b + 5c = 1000 (equation 2)
multiply equation 1 by 2 to get:
4a + 4b + 6c = 1000 (equation 1 multiplied by 2)
4a + 5b + 5c = 1000 (equation 2)
subtract equation 2 from 1 to get:
4a + 4b + 6c = 1000 (equation 1 multiplied by 2)
minus: 4a + 5b + 5c = 1000 (equation 2)
equals: 0 - 1b + 1c = 0
this means that b = c
if b = c, we can solve the original equations again by substituting b for c
to get:
2a + 5b = 500 (equation 1 substituting b for c)
4a + 10b = 1000 (equation 2 substituting b for c)
we multiply equation 1 by 2 to get:
4a + 10b = 1000 (equation 1 multiplied by 2)
4a + 10b = 1000 (equation 2)
we subtract equation 2 from equation 1 to get:
0 + 0 = 0
since that is an identity equation, it means that any value of a or b or c that satisfies any of the two equations for raw material and hours of production will be good for all equations, as long as b = c.
since the profit for c is the greatest, and there is no restriction on how many of "a" need to be produced, the conclusion is that maximizing c will generate the most profit.
the only restriction is that b = c.
for example:
the first equation is:
2a + 2b + 3c = 500
let a = 0
this makes 2b + 3c = 500
this makes b and c each equal to 100 because:
2*100 + 3*100 = 200 + 300 = 500.
with b and c each equal to 100 and a equal to 0, the second equation becomes:
4*0 + 5*100 + 5*100 = 500 + 500 = 1000
the profit will be:
600*100 + 1000*100 = $160,000
any value of "a" greater than 0 will yield a profit that is less than this.
example:
let a = 5
this makes b and c = 500-10 = 490/5 = 98
this was taken from the first equation that said that 2a + 2b + 3c = 500
since a = 5, then 2a = 10
since b = c, then 2b+3c = 5d where d = b = c.
equation becomes 10 + 5d = 500
solving for d gets 5d = 490 which results in d = 98 which means that b and c each equal 98.
profit equation is:
500*5 + 600*98 + 1000*98 = 159300 which is less than 160000.
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this was difficult to solve because a lot of things disappeared and other things became equal to each other.
since a, b, and c could be anything as long as they came up positive and b equaled c, it was difficult to see how to maximize the profit because there was no equation to solve to prove that.
some logic was necessary to deduce that "a" was the weak link in the profit equation and that "a" could be eliminated from the supply side equation with no negative impacts on that because "a" was free to be anything it wanted to be as long as b = c and the raw material and production hours equations were satisfied.
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