Min max: optimal quantity of medicine

  • Thread starter Thread starter Karol
  • Start date Start date
  • Tags Tags
    Max Medicine
Click For Summary
The discussion focuses on finding the optimal quantity of medicine by analyzing the first derivative of the revenue function, R(D). It emphasizes that the maximum of R' occurs when the first derivative equals zero, indicating no change in R with respect to D. Participants clarify that the goal is to maximize R', not R itself, as the maximum point of R corresponds to a flat tangent where small changes in D do not affect R. Suggestions are made to gain a better understanding by substituting numerical values for C and plotting R(D) and R'(D). The conversation highlights the importance of correctly interpreting derivative functions in optimization problems.
Karol
Messages
1,380
Reaction score
22

Homework Statement


Capture.JPG


Homework Equations


Minimum/Maximum occurs when the first derivative=0

The Attempt at a Solution


$$R'=2D\left( \frac{C}{2}-\frac{D}{3} \right)-\frac{1}{3}D^2$$
$$R'=0~\rightarrow~D=C$$
 

Attachments

  • Capture.JPG
    Capture.JPG
    24.6 KB · Views: 773
Last edited:
Physics news on Phys.org
Karol said:

Homework Statement


View attachment 230882

Homework Equations


Minimum/Maximum occurs when the first derivative=0

The Attempt at a Solution


$$R'=2D\left( \frac{C}{2}-\frac{D}{3} \right)-\frac{1}{3}D^2$$
$$R'=0~\rightarrow~D=C$$
Read carefully.
upload_2018-9-19_15-45-13.png


It discusses the point where ##\ R^\prime (D) \ ## is a maximum, not about the max of ## R(D)\,.##

Also, it may help to write R(D) as: ##\ \displaystyle R(D) = \frac{D^2C}{2}-\frac{D^3}{3} \,.##

.
 

Attachments

  • upload_2018-9-19_15-45-13.png
    upload_2018-9-19_15-45-13.png
    9.6 KB · Views: 299
$$R'=2D\left( \frac{C}{2}-\frac{D}{3} \right)-\frac{1}{3}D^2,~~R''=C-2D,~~R''=0:~D=\frac{C}{2}$$
But the greatest change in R for a small change in D is where R has a maximum, hence where R'=0, not where R''=0
 
Karol said:
$$R'=2D\left( \frac{C}{2}-\frac{D}{3} \right)-\frac{1}{3}D^2,~~R''=C-2D,~~R''=0:~D=\frac{C}{2}$$
But the greatest change in R for a small change in D is where R has a maximum, hence where R'=0, not where R''=0
Read the problem again. It's R'(D) which you need to find the maximum for, not finding the maximum for R(D) .
 
Yes, that's correct, i need the maximum for R', but why?
At the point where R has a maximum, i think, a small change in D makes a big change in R
 
Karol said:
Yes, that's correct, i need the maximum for R', but why?
At the point where R has a maximum, i think, a small change in D makes a big change in R

No, that's where R' is a maximum. R' = 0 is where there is no change with a change in D.
 
Karol said:
Yes, that's correct, i need the maximum for R', but why?
At the point where R has a maximum, i think, a small change in D makes a big change in R

No: at the maximum a small change in ##D## makes NO change in ##R##! The tangent line to the graph ##R = f(D)## is horizontal at an interior maximum; that is why we look for points where the derivative vanishes when maximizing or minimizing.

You really need to have a better intuitive understanding of this material, and to help with that I suggest that you substitute some numerical value for ##C##, then plot ##R(D)## and ##R'(D)##.
 
  • Like
Likes SammyS
Thanks
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
Replies
6
Views
1K
Replies
4
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K