SOLUTION: Optimization The power absorbed by a circuit is p(x) = ((V^2)x)/(R+x)^2 where R is the resistance in ohms and V is the voltage in volts. Find the value of x that maximizes the p

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson  -> Lesson -> SOLUTION: Optimization The power absorbed by a circuit is p(x) = ((V^2)x)/(R+x)^2 where R is the resistance in ohms and V is the voltage in volts. Find the value of x that maximizes the p      Log On


   

Question 1030357: Optimization
The power absorbed by a circuit is p(x) = ((V^2)x)/(R+x)^2 where R is the resistance in ohms and V is the voltage in volts. Find the value of x that maximizes the power absorbed.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Find the derivative, I'm substituting a and b for the constants for better readability,
p=%28ax%29%2F%28b%2Bx%29%5E2
dp%2Fdx=%28%28b%2Bx%29%5E2%2Aa-2ax%28b%2Bx%29%29%2F%28b%2Bx%29%5E4
dp%2Fdx=%28a%2F%28b%2Bx%29%5E4%29%28b%5E2%2B2bx%2Bx%5E2-2bx-2x%5E2%29
dp%2Fdx=%28a%2F%28b%2Bx%29%5E4%29%28b%5E2-x%5E2%29
dp%2Fdx=%28a%2F%28b%2Bx%29%5E4%29%28b-x%29%28b%2Bx%29
dp%2Fdx=%28a%28b-x%29%29%2F%28b%2Bx%29%5E3
dp%2Fdx=%28V%5E2%28R-x%29%29%2F%28R%2Bx%29%5E3
So then when the derivative equals zero,
x=R
and
p=%28V%5E2%2AR%29%2F%28R%2BR%29%5E2
p=%28V%5E2%2AR%29%2F%284R%5E2%29
p=V%5E2%2F%284R%29