How to Use Gaussian Elimination to Solve for Vitamin Pills

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Homework Help Overview

The problem involves determining the number of vitamin pills from three brands to meet specific daily requirements for vitamins A, B, and C. The context is set within the framework of Gaussian elimination to solve a system of equations represented by an augmented matrix.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the formulation of an augmented matrix based on the vitamin requirements and the contents of each brand. There are attempts to manipulate the matrix using Gaussian elimination, with some participants expressing confusion over negative values and the implications of infinite solutions.

Discussion Status

There is an ongoing exploration of the relationships between the variables representing the number of pills from each brand. Some participants have provided guidance on defining variables and setting up equations, while others are questioning the validity of their results and the implications of their findings.

Contextual Notes

Participants note constraints related to the requirement for nonnegative whole numbers of pills and the implications of having three unknowns in two equations. There is also discussion about the limits on the values for the number of pills from brand 3 based on the equations derived from the matrix.

  • #31
sdoug041 said:
Ok I see.

If b3 >= 0 , and say b3 was indeed 0, then b2 would have to be -3 to satisfy the equation. Does this mean b3 should be 3<=b3<=8, and 0<=b1<= 5?

Yes, yes, yes. What about b2? Now how many total solutions are there to your pill taking needs?
 
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  • #32
memomator said:
The question states. Find all combination of pills that provide you with the exact daily requirement(no partial pill). How will I find all the combination of pills, knowing only the number of pills in brand 1, 2, and 3.

"the number of pills in brand 1, 2, and 3" is what you need to specify a "combination of pills". For example, from your equations I see 3 pills of brand 3, 5 pills of brand 1 and 0 pills of brand 2 is one solution. How many more are there?
 
  • #33
Ok so 0<=b2<=5

I've found 6 possible solutions for the equations :D

Thanks a lot, Dick.
 
  • #34
Ok i understand that now, is there an easier way to find out how many solutions there are other than plugging into numbers.
 
  • #35
memomator said:
Ok i understand that now, is there an easier way to find out how many solutions there are other than plugging into numbers.

Sure, b1+b3=8 tells you you need b3<=8 to make b1 nonnegative. b2-b3=(-3) tells you you need b3>=3 to make b2 nonnegative. Put them together and you have 3<=b3<=8 with the other values being b1=8-b3 and b2=(-3)+b3. You could also write the answer that way instead of spelling out the explicit values of b1 and b2.
 
  • #36
Thank you very much Dick, I finally understand. You have 6 solutions. Because B3 is between 3 and 8 and the numbers between 3 and 8 are solutions and you take those number and sub them into the other equation to find out b1 and b2.
 

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