# Calculus - Max/Min

1. Apr 1, 2008

1. The problem statement, all variables and given/known data
The electric power in Watts produced is given by p= 144r/(r+.8)^2
where r is the resistance in ohms.
For what value of r is the power P a maximum?

2. Relevant equations

3. The attempt at a solution
Using the quotient rule i found a derivative of..

dp/dt= 144(-r+.8) / (r + .8)^3

Now I need to set this equal to zero and solve for R?

Multiplying the demonminator by 0 would leave me with just the numerator giving me..
-144r + 115.2 = 0

Solving for r gives me r = 0.8ohms

Was there maybe a problem with my derivative?

2. Apr 1, 2008

### rocomath

You did it right, find where the derivative is equal to 0 ... $$144(.8-r)=0$$

Solve for r which you did! Is that not the correct answer?

Last edited: Apr 2, 2008
3. Apr 2, 2008

### Snazzy

Looks right to me. You can always check to see if you've found a local maximum using the first or second derivative tests and plugging it back into the original function.

4. Apr 2, 2008

### rocomath

Test values before and after your max-value: positive to negative should indicate a max-point.

5. Apr 2, 2008

Thanks for confirming it for me. Plugging .8 into the derivative gives me 0 so i guess i was right all along.

6. Apr 2, 2008

### Snazzy

Actually, you'd want to plug in something less than 0.8 and more than 0.8 in the derivative to see if it changes from positive to negative in order for 0.8 to be a maximum.

7. Apr 2, 2008

### jimvoit

Plot the power function for vallues of r from 0 to 10 using a
computer algebra system like Mathematica and you will verify that
the answer 0.8 for the maximum value looks correct.

8. Apr 2, 2008

### rocomath

You don't need to do all that! Just do what Snazzy or I suggested. Make good use of your Algebra skills!

9. Apr 2, 2008