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 =
My 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
can you please confirm this is correct. Thanks
no, maximum power is P(2.8947) = 159.211
ok. I understand that you took the value and plugged in for i. But can I ask why...
because they asked for maximum power, not the current which produces maximum power. Just because you arrive at an answer, that does not mean you have answered the question.
Ok.... great. thank you Steve. I know understand!
I'm sorry, but the calculation you provided seems to be incorrect. Let's go through the process of finding the maximum power step by step.
To determine the maximum power that can be delivered, we need to find the maximum point on the graph of the power function. This can be done by finding the critical points of the function and then evaluating the function at those points.
Given the power function:
P = 110i - 19i^2
To find the critical points, we need to take the derivative of the function with respect to i and set it equal to zero:
P' = 110 - 38i
Setting P' equal to zero, we have:
0 = 110 - 38i
Now, we can solve for i:
38i = 110
i = 110/38
i ≈ 2.8947
So, the critical point occurs at i ≈ 2.8947.
To determine whether this point is a maximum or minimum, we can take the second derivative:
P" = -38
Since the second derivative is negative, the critical point represents a maximum.
Therefore, the maximum power that can be delivered is the power at i ≈ 2.8947. Let's plug this value back into the power equation:
P = 110i - 19i^2
P ≈ 110 * 2.8947 - 19 * (2.8947)^2
Calculating this, we find that the maximum power is approximately:
Max Power ≈ 82.8299 watts
So, the correct answer is approximately 82.8299 watts.