Question 1082236
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.