Question 1082236: A candy manufacturer has 130 pounds of Chocolate-covered cherries and 170 pounds of chocolate-covered mints in stock. He decides to sell them in the form of two different mixtures. One mixtures will contain half cherries and half mints by weights and will sell for $2.00 per pound. The other mixture will contain one-third cherries and two-thirds mints by weight and will sell for $1.25 per pound. How many pounds of each mixture should the candy manufacturer prepare in order to maximize his sales revenue?
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! There's a way to do a quick check on a problem like this:
1/2 c. and 1/2 m. (1:1 ratio) sells for $2/lb
1/3 c. and 2/3 m. (1:2 ratio) sells for 1.25/lb
Right away you know you have to maximize the first mixture, at $2/lb. At ratio 1:1, you get 130 + 130 = 260 of the first mixture. The price will be $520 for this lot. So, is it worth making the second mixture at all? Try it, the answer is no.
:
OK, now let's look at solving this linear problem properly:
Let the mixture of 1/2 cherries and 1/2 mints be A
Let the mixture of 1/3 cherries and 2/3 mints be B
Let x be the number of pounds of A to go in the mix
Let y be the number of pounds of B to go in the mix
:
The revenue function can be expressed as:
z = 2x+1.25y
Now, since each pound of A contains 1/2 lb of cherries and each pound of B contains 1/3 pound of cherries, the total number of pounds of cherries used in both mixtures is:
1/2x + 1/3y
And the pounds of mint:
1/2x + 2/3y
Limits:
- There are 130 lbs of cherries
- There are 170 lbs of mints
Therefore:
1/2x + 1/3y <= 130
1/2x + 2/3y <= 170
For our calculations, x >= 0 and y >= 0
So now we can formulate the problem:
Find x and y that maximize z = 2x + 1.25y subject to these constraints:
1/2x + 1/3y <= 130
1/2x + 2/3y <= 170
x >= 0
y >= 0
Evaluate the objective function of each of the points and find that the candy manufacturer attains maximum sales of $520 when he produces 260 pounds of mixture A and none of mixture B.
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