Restrictions on variables.

1. Apr 20, 2016

Kirito123

1. The problem statement, all variables and given/known data
Jeffrey is planning a large dinner party. His budget is $3000. He has the option of serving chicken at$30 per meal or roast beef at $40 per meal. The cost of the party can be modeled with the equation 30c + 40b = 3000, where c represents the number of chicken meals ordered and b represents the number of roast beef meals ordered. What are the restrictions on the variables? Justify your reasoning. 2. Relevant equations restrictions on variables. 3. The attempt at a solution Well for the variable c you cannot exceed 100 meals or you will exceed your budget and you cannot have – 1 or more meals/$. So it has to be 0 or greater. So you can say:

0 <_ C <_ 3000 ($) (Note: the <_ is meant to represent greater or equal to.) Same with b you cannot exceed 75 beef meals or you u will exceed your budget. You also can’t have –1 meals or dollars. So we can say that it has to be 0 or greater. 0 <_ B <_ 3000 ($)

Its a new concept for me and I would like to know if I am understanding it correctly, and got the answer right. I appreciate your time and effort and would like to say thank you in advance.

2. Apr 20, 2016

Ray Vickson

Are you sure you have presented the problem correctly? There are two possibilities, (a) and (b) below.

(a) If Jeffrey can spend at most $3000, then the constraint should be $30 c + 40 b \leq 3000$, and, of course, $b, c \geq 0$. (b) If Jeffrey must spend exactly$3000, then the constraint is as you have written it: $30 c + 40 b = 3000$, and, of course, $b,c \geq 0$.

3. Apr 20, 2016

Kirito123

Ok I think I get what your saying, Jeffrey doesn't have to spend exactly $3000. He just can't spend more then that budget. So im guessing that the answer (a) that you suggested is right, since Jeffrey can only spend at most$3000. Also how would I explain that in steps??

4. Apr 20, 2016

Kirito123

i dont understand what to do? is my answer correct?

5. Apr 20, 2016

Kirito123

I came up with another answer: please tell me if this is correct!
Assuming that Jeffrey is purchasing both chicken and roast beef, but only has a budget of $3000, the budget should be divided amongst both chicken and beef meals. For chicken meals, it cannot exceed 50 meals, or this would surpass$1500 dollars. And for beef it cannot exceed around 37 meals, or this would exceed $1500. Restriction for chicken meals: 0 < C < 1500 ($)

Restriction for beef meals: 0 < B < 1500($) Therefore the constraint should be: 30c + 40b < 3000 and b, c > 0 6. Apr 23, 2016 haruspex In your original post, you had 30c+40b=3000. It reads as though you were given that equation as part of the problem specification. Is that so, or did you make up that equation based on the$3000 budget?

7. Apr 23, 2016

If, in fact, there are additional restrictions on Jeffrey's spending patterns (such as a $1500 limit on each of beef and chicken) then those must be stated as part of the problem's description. It is not valid to simply add them later for no really good reason. 9. Apr 23, 2016 Kirito123 I see. 10. Apr 23, 2016 Kirito123 So basically since they did not give us like you have to spend this much on beef or chicken etc. then my formula i mentioned was wrong. But since i can spend only up to 3000 dollars on beef or chicken that's why its 30c + 40b < 3000. that means both chicken and beef can not pass 3000 in total. correct? 11. Apr 23, 2016 Kirito123 what do you mean by that? 12. Apr 23, 2016 haruspex Sorry, forget it. I misread your post. Too hasty. 13. Apr 23, 2016 Kirito123 Ok lol 14. Apr 23, 2016 Ray Vickson Yes, we have already said that. That means that 30c + 40b <= 3000, NOT "< 3000". If you said "< 3000" you would be saying "<= 2999.99", so you would be allowing Jeffrey to spend any amount up to$2999.99, but he would not be allowed to spend one more penny to bring it up to \$3000.

15. Apr 23, 2016

Kirito123

Ok i got it thx for the help :)

16. Apr 24, 2016

ehild

I do not see the other restriction that both b and c must be positive integers.

17. Apr 24, 2016

Ray Vickson

Well, non-negative integers at least.