How many boxes of each mixture should the company make to maximize profit? Linear Programming Problem

Question

A company makes two snack mixtures. A box of mixture A contains 6 ounces of peanuts, 1 ounce of M&M's, and 4 ounces of raisins and sells for \$4.25. A box of mixture B contains 12 ounces of peanuts, 3 ounces of M&M's, and 2 ounces of raisins and sells for \$6.55. The company has available 5400 ounces of peanuts, 1200 ounces of M&M's, and 2400 ounces of raisins. How many boxes of each mixture should the company make to maximize profit?

I'm suppose to solve using a graph, and I've been watching youtube videos for now over 3 hours and I'm feeling very stuck. I started off wrong because I was trying to do a equation for 2 variables because I thought box A and box B would be 2 variables but now I'm thinking the food would actually make it 3 variables meaning I just did all that work for nothing. I would really appreciate some further assistance on how you guys found the answer. My answer was $x=500$, $y=200$, $z(op)=3685$ .... please tell me I didn't completely mess this up. Thanks!

Answer 1

To do it by hand, imagine you only make A boxes. How many can you make? What is the limiting component? How much money do you make? Now if you make fewer A boxes and free up some of the limiting component, allowing you to make B boxes, is that good? For example, if peanuts were limiting you would have to skip two boxes of A to make a box of B. That is a bad deal because you trade two 4.25s for one 6.55. If the trade is profitable, keep doing it until you run out of something.

Answer 2

What are you trying to maximize or minimize? Ostensibly, the profit. This would be $$ 4.25A + 6.55B $$

So yes, there are two variables here.

How many boxes can you make?

Well, that depends on your supplies. So the three food items will create three constraints. For example, If each A box uses 6 oz. peanuts and each B box uses 12 oz. peanuts, then $6A + 12B$ better be less than or equal to 5400, unless you can magically create some peanuts!

And yes, I get the same breakup that you do, but $500A + 200B \neq 3685$.

Answer 3

Formulation of the LPP:

Step 1: Declare the decision variables. You should decide how many $A$ and $B$ to produce. Let $x$ and $y$ be the numbers of the boxes, respectively.

Step 2: Make up the objective function. The profit (or revenue) function is $R=4.25x+6.55y$.

Step 3: Define the constrains. $$\begin{cases}6x+12y\le 5400 \ \ \text{(peanuts constraint)}\\ \ \ \ \ x+3y\le 1200 \ \ \text{(M&Ms constraint)}\\ \ \ 4x+2y\le 2400 \ \ \text{(raisins constraint)}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ x\ge 0 \ \ \ \ \ \ \ \ \text{(non-negativity constraint)}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ y\ge 0 \ \ \ \ \ \ \ \ \text{(non-negativity constraint)}\\ \end{cases}$$

Solution of the LPP (graphical):

Step 1: Draw the feasible region (the green area):

$\hspace{2cm}$

Step 2: Find the corner points of the feasible region: $$O(0,0); A(0,400); B(300,300); C(500,200); D(600,0).$$

Step 3: Evaluate the objective function at the corner points and choose the optimal value:

$$\begin{align}R(0,0)&=0;\\ R(0,400)&=2620;\\ R(300,300)&=3240;\\ R(500,200)&=3435 \ \text{(max)};\\ R(600,0)&=2550.\end{align}$$ See WA solution.