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November 26, 2014

November 26, 2014

Posted by **Shirley** on Saturday, February 23, 2008 at 8:42pm.

Find the maximum power (in watts)that can be delivered in this circuit.

I had posted this question before, and someone gave me this solution:

- 8 i^2 + 240 i = P

i^2 - 30 i = - P/8

i^2 - 30 i +(30/2)^2 = -P/8 + 225

(i-15)^2 = -P/8 + 225

vertex at i = 15 and P at 225*8 = 1800

From this part on, I don't really get it:

i^2 - 30 i +(30/2)^2 = -P/8 + 225

(i-15)^2 = -P/8 + 225

vertex at i = 15 and P at 225*8 = 1800

Can anyone please explain it to me???THANKS A LOT!

- math -
**drwls**, Sunday, February 24, 2008 at 9:44amThere are two ways to do this kind of problem. One is to use calculus, which leaves you with the equation

2I - 30 = 0, leading right away to I = 15. The other way, which does not require calculus, is to use the method of "completing the square". That is what was done in the example you posted. When the equation

P = 240I-8I^2 is rewritten

I^2 - 30 I +(30/2)^2 = -P/8 + 225

= (I-15)^2 , or

P/8 = 225 -(I - 15)^2

you see right away that the maximum value of P is obtained when I = 15.

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