Linear Programming(Max. and Min. of a function)

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sillymoiksta

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I've had trouble trying to figure out this problem and a few others very similar to it. I am still very confused about how to do these types of problems. I've asked my dad and oldest brother but they are confused by it as well. The question is:

A sporting goods manufacturer makes a $5 profit on soccer balls and a $4 profit on volleyballs. Cutting requires 2 hours to make 75 soccer balls and 3 hours to make 60 volleyballs. Sewing needs 3 hours to make 75 soccer balls and 2 hours to make 60 volleyballs. Cutting has 500 hours available, and Sewing has 450 hours available.
a. How many soccer balls and volleyballs should be made to maximize the profit?
b. What is the maximum profit the company can make from these two products?

I think I need to make 4 inequalities for this problem. I wasn't sure whether I should define my variables by hours or by # balls made. I wrote down s=#soccer balls and v= # volleyballs but I wasn't sure if that was right. Then I wrote s>=0 and v>=0 as my first two inequalities. The other two should involve the time spent cutting and sewing, but I'm not sure how to write that inequality other than it will end in <=500(less than or equal to) for cutting and <=450 for sewing. Once I get my inequalities I can graph them and find the vertices of the feasible region. And from there I can find the maximum and minimum. So what I really need to know is how do you get the rest of the inequalities? And what should the variables be defined as?
I'd appreciate any help anyone can offer. Thank you very much! :D
 
Write Definitions.

S = # of Soccer Balls to Produce
V = # of Volley Balls to Produce

Profit(S,V) = 5*S + 4*V <== This is what you wish to maximize.

Cutting:

Note: This information is a little weird because it does not give you the unit cost. You need to make it unit cose.

(2/75)*S + (3/60)*V <= 500 <== Make sure you get the units right. (hrs/ball)*ball = hrs

Sewing:

(3/75)*S + (2/60)*V <= 450

End Note: It MAY be the case that the manufacturer is willing to make soccer balls ONLY in lots of 75 and volley balls ONLY in lots of 60. In this case, you can run the whole thing by LOT, rather than by individual ball. It will look a little different.

S = # of 75-Lots of Soccer Balls to Produce
V = # of 60-Lots of Volley Balls to Produce
Profit(S,V) = 5*(75*S) + 4*(60*V)
2*S + 3*V <= 500
3*S + 2*V <= 450
 
Thank you so much for your help! Thanks for being so detailed too and explaining everything so nicely. I hadn't thought about it being sold as lots or individual, but I think I'm supposed to solve it as a lot. I'm not sure because the problem doesn't say, but that's the one I ended up solving for.
Using the lot equations I got a max of f(150,0) =56,250, meaning 150 soccer balls and 0 volleyballs must be sold to get a max value of $56, 250. I hope I did that right. Thanks again for your wonderful help! :D I was so happy to see your reply to my question!
 
Only one follow-up question, if you don't mind.

On your last equation, you had it as

3*S + 60*V <= 450

Did you mean to say 3*S + 2V <= 450 ?

Thanks again
 
Ah. Thanks fo rthe edit. It is repaired above.

"Lots" makes this a little bit trickier, as you cannot get an answer such as 83 soccer and 62 volley. they must come out multiples of 75 and 60, as you seem to have noticed.

Rounding only to integer balls, rather than integer lots:
The critical points are (Soccer, Volley), (0,500/3)=(0,166), (150,0), and (70,120)
Rounding to lots gives:

(0,166) ==> (0,120) a huge loss (44 balls from the frontier solution) to volley ball production
(150,0) ==> (150,0) hitting a lot right on is VERY encouraging.
(70,120) ==> (0,120) discarding the soccer balls entirely cannot be a good thing. Not to mention we already foundf this one.
Other possibilities might be (75,0), (75,60), and (0,60). Since (75,0) and (0,60) are very obviously not advantageous, (75,60) is the only one that merits attention.

Having to hit integer multiples makes the task a bit harder, as the algebra doesn;t necessarily know where the solutions are hiding.

As for your maximum profit, let's review: $56,250 / 150 = $375 - You're selling soccer balls for over $375!!?? Wow! That must be some soccer ball! Rethink that profit equation a bit. Look very carefully at the units. Are they "per ball" or "per lot"?
 
No, no! These are just magic profit-making balls.

Indeed, it is hoped that this is an arithmetic exercise and not a business plan.
 
Thanks again for all your help. My teacher gave us the 4 equations he wanted us to use today as he realized that this problem was tricky. He said to use s>=0, v>=0, 2/75s + 3/60v <= 500, 3/75s + 2/60v <=450. So I take it that we're now solving for individual soccer and volleyballs now. I graphed those 4 equations and got a triangle shaded region with vertices of (0, 11,250), (0,0), and (11,500, 0). My max profit was f(11,500, 0) = 57, 500.
But is that incorrect because you said before when I got answer that high that that wouldn't make sense. I don't see what I did wrong but that answer still seems high. Any thoughts?
Thanks again so much. I really appreciate it because I want to get better at doing these types of problems. There might even be one like this on the quiz tomorrow.
 
No worries. Check your work and make sure. Doing individual balls has excellent potential for more profit. Whacking off production because we are stuck with a lot size is a common industry waste.
 
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