Max/Min Calculus Homework: Find Resistance for Max Power

[uradnky]

Homework Statement

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?

Homework Equations

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?

 

[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?

 

[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.

 

[rocomath]

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

 

[uradnky]

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

 

[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.

 

[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.

 

[rocomath]

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.

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

 

[uradnky]

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.

Yeah, with 1st derivative test right?
I just figured that it would be ok to assume a maximum because that's what the question asked for, and solving for r only gave me one value.