Cfrgtkky

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!

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.
$$begin{cases}6x+12yle 5400 text{(peanuts constraint)}\
x+3yle 1200 text{(M&Ms constraint)}\
4x+2yle 2400 text{(raisins constraint)}\
xge 0 text{(non-negativity constraint)}\
yge 0 text{(non-negativity constraint)}\
end{cases}$$

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.

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.

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 $500A + 200B neq 3685$.