How to minimize cost with two different sizes of drug vials?

  • Context: Undergrad 
  • Thread starter Thread starter sub25
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on minimizing costs when administering a total drug dose using two vial sizes: 300mg vials costing $400 each and 500mg vials costing $600 each. For a required dose of 1100mg, the optimal solution is to use 2 vials of 300mg and 1 vial of 500mg. The cost function is defined as Total Cost = 400x + 600y, subject to the constraint 300x + 500y ≥ 1100mg. The recommended approach includes solving the equation for y in terms of x and minimizing the resulting quadratic function.

PREREQUISITES
  • Understanding of linear equations and inequalities
  • Familiarity with quadratic functions and their minimization
  • Basic knowledge of programming for optimization (optional)
  • Experience with cost analysis in pharmaceutical contexts
NEXT STEPS
  • Learn how to solve linear programming problems using the Simplex method
  • Explore quadratic function minimization techniques, including completing the square
  • Investigate programming solutions for optimization problems using Python or R
  • Study real-world applications of cost minimization in pharmaceutical pricing strategies
USEFUL FOR

Pharmaceutical analysts, cost optimization specialists, and anyone involved in drug dosage calculations and pricing strategies will benefit from this discussion.

sub25
Messages
1
Reaction score
0
There is a drug that comes in two size. 1 is 300mg and cost 400 and 1 is 500mg and cost 600. Given a total dose needed, how many of each vile type do I need to minimize the cost.

Ex) I need a 1100mg dose, how many 300mg viles and how many 500 viles do I need to minimize the cost. I know the answer from looking at it is 2-300mg and 1-500mg.

So if x = # of 300mg viles and y = # of 500mg viles we have :

Total Cost = 400 * x + 600 * y; I also have the constraint that ( for this paticular example) 300x + 500y >= 1100mg.

I feel like this should be easy but my mind is not working.

Thanks

-Marc
 
Physics news on Phys.org
Better to write 300x+ 500y= 1100. Then if x and y are not integers, choose the next larger integer.

Solve 300x+ 500y= 1100 for y, say, as a function of x. Replace y in the cost equation by that. That will result in a quadratic function of x which you can minimize by completeing the square.
 
In some cases (but not in this one), the conversion from discrete to continuous isn't so forgiving. If this were one such case, you might do better to just write a program to check all possibilities and keep the largest. i.e.

Code:
for every point (x,y) in the domain
   do see if this yields the best solution so far
      if so, save this.
the optimum is the best solution so far.

For the record, you should usually use HallsOfIvy's method, except for trick problems designed to mess with your head.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
14
Views
3K
Replies
1
Views
3K
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
7
Views
3K
Replies
5
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K