Linear Programming

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tootsiejt714

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Sep 19, 2010
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Hello there,

I would appreciate your help on the following problem. This is a new topic to me and I am running into difficulty with the following problem and how to phrase the constraints of the problem in a mathematical equation. The whole problem is as follows:

The Cheddar Cheese Company produces two cheese spreads: Regular and Zesty. The cheese spreads are packaged in 12-ounce containers, which are then sold to distributors throughout the country. The Regular spread contains 80% mild cheddar and 20% extra sharp cheddar, and the Zesty spread contains 60% mild cheddar and 40% extra sharp cheddar. This year, a local dairy cooperative offered to provide up to 8100 pounds of mild cheddar cheese and up to 3000 pounds of extra sharp cheddar ( 1 pound = 16 ounces ). Each container of Regular spread is sold for a profit of $1.95 and each container of Zesty spread is sold for a profit of $2.20. Formulate a linear programming model that can be used to determine the number of containers of Regular and Zesty spreads to produce in order to maximize profit.

So far this is what I have been able to come up with:

Xr = # of containers of Regular spread to produce
Xz = # of containers of Zesty spread to produce

Max 1.95Xr + 2.20Xz

s.t. (constraints)
Xr>=0
Xz>=0
.80Xr + .60Xz <=8100 (mild cheddar constraint) ???????
.20Xr + .40Xz <=3000 (extra sharp cheddar constraint) ???????

I know there should be more constraints and I believe there should be some conversions involved with the ounce vs. pound, but I am not sure how to approach it???? Can you please confirm what I have so far is correct or incorrect and please provice your help.

Thanks.
 


Yes, you need to convert the two given weights in pounds to ounces because the percentages of relative cheese-type are all ounces. (Or, alternatively, you could leave them in pounds, and convert the 12-ounce containers to equivalent pound-containers. Your choice.)

Also, look at your inequality: 0.8 Xr + 0.6 Xz <= 8100

This says, "80% of the number of "regular" containers PLUS 60% of the number of "zesty" containers add up to 8100 pounds or less".

Numbers of containers do not add up to pounds. Adding numbers of containers together always results in another number of containers, not a weight.

The inequality needs to say, "80% of the 12 ounces of product in all of the "regular" containers PLUS 60% of the 12-ounces of product in all of the number of "zesty" containers add up to 129600 ounces or less".

In both of your equations with percents, the percentages come from total ounces of cheese-type, not from the number of containers.

I mean, multiply the number of containers by 12 ounces to get the total number of ounces of product, and then take a percent of that, in each of those two inequalities.

0.8(12 Xr) + 0.6(12 Xz) <= 129600

The graphs of the two inequalities in Quadrant I (also including the positive Xr and Xz axes) form a shape. Determine the coordinates at each vertex of this shape (ignoring the Origin), and use them to evaluate the profit expression:

P = 1.95 Xr + 2.2 Xz

You'll see which (Xr, Xz) pair produces the greatest P.

 
You should get Regular = 9600 and Zesty = 5200, for revenue of $30,160
 
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