Inequality Word Problem: donuts, bagels, and min. costs

[meg2001]

Your homeroom has invited another homeroom over for breakfast and you are in charge of bringing food. Both groups voted, and there was a tie between donuts and bagels; you decided that in order to make everyone happy it would be possible to bring both. You need food for at least 44 hungry students. A box of a dozen donuts feeds 8, while a bag of a dozen bagels is enough for 12 students. Based on the results of the votes, it is clear that you need at least 18 donuts and at least 12 bagels. A dozen donuts will cost $5.50, while a dozen bagels is $7.75. How many donuts and bagels should you buy in order to minimize your cost? What is the total cost? Assume that you can’t buy either donuts or bagels in packages of less than a dozen.

So far for restrictions i have x+y?44 Im not sure where else to go with this problem Plz help thanks!

 

[stapel]

So far for restrictions i have x+y?44 Im not sure where else to go with this problem

How have you defined "x" and "y"? What do they stand for?

What, in words, are the limitations and requirements that must be fulfilled?

Thank you!

Eliz.

 

[meg2001]

x=donut y=bagels
k heres the limitations i think for the problem
x?18 y?12
and heres the equation for finding the minimum cost 5.5x+7.75y=Cost
thats basically what i know how to set up so far but this one confuses me "A box of a dozen donuts feeds 8, while a bag of a dozen bagels is enough for 12 students." I dont know how to express that sentence in math.

 

[meg2001]

ok i think i solved the problem after some more work i got 1 dozen bagels which feeds 12 and then i got 4 dozen donuts which feeds 32. So my minimum cost in the end comes out to $29.75 Is that correct and did i get all my limitations down from the previous post?

 

[Deleted member 4993]

x=donut y=bagels
k heres the limitations i think for the problem
x?18 y?12
and heres the equation for finding the minimum cost 5.5x+7.75y=Cost<<< Those prices for dozens - not single

so the equation should be

5.5 * x/12 + 7.74 * y/12 = cost

and

x/12 * 8 + y/12 * 12 >= 44

thats basically what i know how to set up so far but this one confuses me "
A box of a dozen donuts feeds 8, while a bag of a dozen bagels is enough for 12 students." I dont know how to express that sentence in math.

 

[Deleted member 4993]

ok i think i solved the problem after some more work i got 1 dozen bagels which feeds 12 and then i got 4 dozen donuts which feeds 32. So my minimum cost in the end comes out to $29.75 Is that correct and did i get all my limitations down from the previous post?

If you want us to check your work - you have to show your work.

 

[stapel]

x=donut y=bagels

The number of donuts and bagels? The number of boxes of donuts and bagels? The number of dozens of donuts and bagels (by which they are priced)? The amount spent on donuts and bagels? Or something else?

x?18 y?12

Here, you appear to be using "x" and "y" to stand for the actual number of items.

...the equation for finding the minimum cost 5.5x+7.75y=Cost

But here you appear to be using "x" and "y" to stand for the number of dozens of items. So the first step would probably be to pick one fixed definition for your variables (perhaps in terms of dozens? since this is how you will be ordering and how they are priced), and then work from there, noting, for instance, that "18" is 1.5 dozen.

Eliz.

 

[meg2001]

yea i wanted to use x and y representing the dozens so i got screwed up with Khans work seems like your making it too confusing so im just keeping the x and y set to a dozen. For this problem i think its kind of simple all i did was buy 4 dozen donuts which can feed 32 students. So i only need to feed 12 more to meet my goal and the 1 dozen bagels does that and it seems i meet all my goals.

 

[Deleted member 4993]

yea i wanted to use x and y representing the dozens so i got screwed up with Khans work seems like your making it too confusing so im just keeping the x and y set to a dozen. For this problem i think its kind of simple all i did was buy 4 dozen donuts which can feed 32 students. So i only need to feed 12 more to meet my goal and the 1 dozen bagels does that and it seems i meet all my goals.

But how do you that is the minimum cost?

 

[meg2001]

try any other combination and u will see 3 dozen donuts and 2 dozen bagels is more and so is any other that i have tried

plus i graphed it and used y>=44-x and then put in the other two lines y>=12 and found the intersection with that point and that was (32,12)

 

[Deleted member 4993]

try any other combination and u will see 3 dozen donuts and 2 dozen bagels is more and so is any other that i have tried

plus i graphed it and used y>=44-x <<< This is not one of the constraints - if 'x' and 'y' are in dozens.

and then put in the other two lines y>=12 <<< This is not one of the constraints - if 'x' and 'y' are in dozens.


and found the intersection with that point and that was (32,12)

 

[meg2001]

yea what ever too confused with all this i will just talk to my teacher tomorrow was worth a try going online

 

[Deleted member 4993]

Your homeroom has invited another homeroom over for breakfast and you are in charge of bringing food. Both groups voted, and there was a tie between donuts and bagels; you decided that in order to make everyone happy it would be possible to bring both. You need food for at least 44 hungry students. A box of a dozen donuts feeds 8, while a bag of a dozen bagels is enough for 12 students. Based on the results of the votes, it is clear that you need at least 18 donuts and at least 12 bagels. A dozen donuts will cost $5.50, while a dozen bagels is $7.75. How many donuts and bagels should you buy in order to minimize your cost? What is the total cost? Assume that you can’t buy either donuts or bagels in packages of less than a dozen.

So far for restrictions i have x+y?44 Im not sure where else to go with this problem Plz help thanks!

Let

B = dozens of bagels

D = dozens of donuts

then

B >= 1

D >= 1.5

B*12 + D*8 >= 44

cost = 7.74 * B + 5.5*D to be minimized