# Linear Programming Question - Completed just needs checking

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• December 28th 2010, 06:09 PM
Linear Programming Question - Completed just needs checking
Hi,

I have a question I need to answer.

Company A produces and sells a popular pet food product packaged under two brand names, with formulas that contain different proportions of the same ingredients. Company A made this decision so that their national branded product would be differentiated from the private label product. Some product is sold under the company’s nationally advertised brand (Brand X), while the re-proportioned formula is packaged under a private label (Brand Y) and is sold to chain stores.

Because of volume discounts and other stipulations in the sales agreements, the contribution to profit from the Brand Y private label product is only \$30 per case compared to \$40 per case for product sold to distributors under the company’s Brand X national brand.

An ample supply is available of most of the pet food ingredients; however, three additives are in limited supply. The tight supply of nutrient C (one of several nutrient additives), a flavor additive, and a color additive all limit production of both Brand X and Brand Y.

The formula for a case of Brand X calls for 4 units of nutrient C, 12 units of flavor additive, and 6 units of color additive. The Brand Y formula per case requires 4 units of nutrient C, 6 units of flavor additive, and 15 units of color additive. The supply of the three ingredients for each production period is limited to 30 units of nutrient C, 72 units of flavor additive, and 90 units of color additive.

A. Determine the equations for each of the three constraints that are plotted on the attached “Graph 1.”
1. Identify each constraint as a minimum or a maximum constraint.

B. Determine the total contribution to profit that lies on the objective function (profit line) as it is plotted on the graph if the company produces a combination of cases of Brand X and Brand Y.

C. Determine how many cases each of Brand X and of Brand Y you recommend should be produced during each production period for optimum production if Company A wants to generate the greatest amount of profit.

D. Determine the total contribution to profit that would be generated by the production level you recommend in part C.

I have got the answers below:

Objective A
The constraint for nutrient C = 4x+4y is less or equal than 30 which is the minimum constraint. Therefore y<=-x+7.5 is the maximum constraint.
The constraint for flavor additive = 12x+6y is less or equal than72 which is the minimum constraint.. Therefore y<=-2x+12 is the maximum constraint.
The constraint for color additive = 6x+15y is less or equal than 90 which is the minimum constraint.
Therefore y<=-2x/5+6 is the maximum constraint.
Objective B
.Objective function (P) is (P) = \$40x+\$30y
Objective C
The vertices are:
P(0, 0) = \$0 + \$0 = \$0 profit
P(0, 6) = \$0 + \$ 180 = \$180 profit
P(2.5, 5) = \$100 + \$150 = \$250 profit
P(4.5, 3) = \$180 + \$90 = \$270 profit
P(6, 0) = \$240 + \$0 = \$240 profit
Therefore the company should produce 4.5 cases of brand X and 3 cases of brand Y in each optimum period.
Objective D
The total contribution to profit that each period would produce is \$270.

If someone could let me know if I have got these right I would appreciate it. If there are errors could you please point me in the direction to fixing them.

I have attached a copy of the graph.

Thanks,
• December 28th 2010, 06:31 PM
pickslides
I agree with your graph and calculations so far, I think you have forgotten a corner point to test (3.75,4.5)

Also is there any restrictions on the cases themselves? Do the units have to be integers?
• December 28th 2010, 06:41 PM
I wasn't sure about (3.75, 4.5); that is the intersection between color and flavor. So I use every corner point where any of two line intersect?

The smallest size of cases I can use are half cases. So the above point would be okay.
• December 28th 2010, 06:43 PM
pickslides
Quote:

So I use every corner point where any of two line intersect?

(Nod)
• December 28th 2010, 06:52 PM
Quote:

Originally Posted by pickslides
(Nod)

Awesome, well I will submit this after correcting the answers to C & D after finding out about missing point.(Blush)

Thanks,
• December 29th 2010, 07:21 AM
Okay this came back as neededing revision.
A: This was correct except for the minimum constraint.
B: I got the response "The profit function is correct. However, a dollar value should be included as well. Reviewing how to use the profit line to compute this amount will help in revising the work."
C: I got the response "Most of the points included in the analysis are correct. However, one point included is not a critical point on the graph and interferes with the accuracy of the work." I think they are telling me I shouldn't have enter the point that was recommended becayse of the comment in D.
D: "Look for a lower value upon revision of section C."

Any help is appreciated.
• January 4th 2011, 06:29 PM
Quote:

B: "The profit function is correct. However, a dollar value should be included as well. Reviewing how to use the profit line to compute this amount will help in revising the work."

I have worked everything else out except this. I put the answer as Objective function (P) is (P) = \$40x+\$30y and I got the response above. Any help please?
• January 4th 2011, 06:41 PM
pickslides
Substitute the corner point $(x,y)$ which gave a max into this equation. $P=40x+30y=\dots$
• January 4th 2011, 06:55 PM
Quote:

Originally Posted by pickslides
Substitute the corner point $(x,y)$ which gave a max into this equation. $P=40x+30y=\dots$

Looking at the graph in my first post I came up with (P) = (\$40*6) + (\$30*8) = \$480
• January 4th 2011, 07:10 PM
pickslides
I see no corner point at (6,8)

Quote:

P(0, 0) = \$0 + \$0 = \$0 profit
P(0, 6) = \$0 + \$ 180 = \$180 profit
P(2.5, 5) = \$100 + \$150 = \$250 profit
P(4.5, 3) = \$180 + \$90 = \$270 profit
P(6, 0) = \$240 + \$0 = \$240 profit

and from post #2

Quote:

Originally Posted by pickslides

I think you have forgotten a corner point to test (3.75,4.5)

which of these points gives a maximum?
• January 4th 2011, 07:35 PM
Quote:

Originally Posted by pickslides
I see no corner point at (6,8)

and from post #2

which of these points gives a maximum?

It is P(4.5, 3) = \$180 + \$90 = \$270 profit

So answer to B would be P = (\$40*4.5) + (\$30*3) = \$270?
or would I write P = (\$40X) + (\$30Y) = \$270?

I don't think it's that though because I have received this response.
Comment: The value of \$270 is incorrect. Look for a different value than the one presented in section D. Pay attention to the words of and to. A dollar amount must also be provided. Please carefully review the profit line to determine which two separate points can be used to determine this amount. Both points should yield the same amount and zero is an acceptable value for X and/or Y.

That's why I thought it was 6,8 because that's where the dotted line runs on the graph. The underlined part made me think that. So yep confused.
• January 5th 2011, 04:27 PM
Finally worked in out (P) = (\$40*6) + (\$30*0) = \$240.

• December 16th 2012, 12:30 PM
XPLIZITO
Re: Linear Programming Question - Completed just needs checking
Trying to follow your posts but they are a little confusing... can you show the final product with the correct answers so I can follow more clearly?
• December 16th 2012, 12:46 PM
skeeter
Re: Linear Programming Question - Completed just needs checking
Quote:

Originally Posted by XPLIZITO
Trying to follow your posts but they are a little confusing... can you show the final product with the correct answers so I can follow more clearly?

you do know you're commenting on a thread that's almost two years old ... ?
• December 16th 2012, 12:50 PM
XPLIZITO
Re: Linear Programming Question - Completed just needs checking
Yes, I noticed but thought I'd still give it a shot. I've got the same assignment almost completed and would like some guidance in order for it not to be returned for corrections. I really need the help.
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