The power P delivered to a load by a current of i amperes through a resistance of 19 ohms, when 110 volts are supplied at one end of the line, is given by the formula P = 110i - 19i2 . Determine the maximum power that can be delivered.

maximum power in watts =

Your umpteen posts have been removed.

Once you write up YOUR ATTEMPT at each problem, please re-post, and someone here will be happy to comment.

Oh, sorry I didn't realise that I needed to provide my answer. I just want to check to see if I obtain the correct answer.

My answer is:

P=110i-19i^2
P'=110-38i
0=110-38i
i=2.8947
P"=-38 therefore a max

Max Power in watts = 2.8947

Please re-post your questions along with your answers, and a math teacher will check your work.

Also there's no need to switch user names.

Hmmm. I agree with i at 2.89 amps

Powermax=put i in the formula for power given, and you have it.

To determine the maximum power that can be delivered, we need to find the maximum point of the power equation P = 110i - 19i^2.

To find the maximum point, we can take the derivative of the power equation with respect to i and set it equal to zero.

dP/di = 110 - 38i

Setting the derivative equal to zero, we have:

110 - 38i = 0

Solving for i, we get:

38i = 110

i = 110/38
i ≈ 2.895

Now that we have the value of i, we can substitute it back into the power equation to find the maximum power:

P = 110i - 19i^2

P = 110(2.895) - 19(2.895)^2
P ≈ 318.45 - 19(8.377025)
P ≈ 318.45 - 159.57
P ≈ 158.88

Therefore, the maximum power that can be delivered is approximately 158.88 watts.