Max/Min Calculus Homework: Find Resistance for Max Power

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SUMMARY

The maximum power produced by an electric circuit is achieved at a resistance of 0.8 ohms, as derived from the power function p = 144r/(r + 0.8)^2. The derivative of the power function, dp/dr = 144(-r + 0.8)/(r + 0.8)^3, is set to zero to find critical points. The first derivative test confirms that this point is indeed a maximum, as the sign of the derivative changes from positive to negative around r = 0.8. Additionally, plotting the function using a computer algebra system like Mathematica supports this conclusion.

PREREQUISITES
  • Understanding of calculus, specifically the quotient rule for derivatives.
  • Familiarity with the first derivative test for identifying local maxima.
  • Basic knowledge of electric power formulas and resistance in ohms.
  • Experience with computer algebra systems, such as Mathematica, for graphing functions.
NEXT STEPS
  • Learn how to apply the first derivative test in calculus.
  • Explore the use of Mathematica for plotting functions and analyzing critical points.
  • Study the implications of resistance on electric power in circuits.
  • Investigate advanced calculus techniques for optimization problems.
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Students studying calculus, electrical engineering students, and anyone interested in optimizing electric power systems.

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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?
 
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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:
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.
 
Test values before and after your max-value: positive to negative should indicate a max-point.
 
Thanks for confirming it for me. Plugging .8 into the derivative gives me 0 so i guess i was right all along.
 
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.
 
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.
 
jimvoit said:
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!
 
Snazzy said:
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.
 

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