In a 120 volt electrical circuit having a resistance of 16 ohms, the available power P in watts is a function of I, the amount of current flowing in amperes. If P(I)=120I-16I^2, how many amperes will produce the maximum power in the circuit? What is the maximum power?

Calculus:

dP/dI=0=120-32I....I=120/32 amps.
Graphing:
y=-16x^2+120x =x(-16x+120) and those roots (when y=0) (x=0, or x=120/16)
well, it is a parabola, crossing the x axis at those roots, so the max should be half way between the roots, or at x=1/2 (120/16 +0) + O = 120/32

To find the amount of current that will produce the maximum power in the circuit and the corresponding maximum power, we need to find the maximum point of the function P(I) = 120I - 16I^2.

Step 1: Take the derivative of P(I) with respect to I.
P'(I) = 120 - 32I

Step 2: Set the derivative equal to zero and solve for I to find the critical point.
120 - 32I = 0
32I = 120
I = 120/32
I = 3.75 amperes

Step 3: To confirm that this is the maximum point, take the second derivative of P(I) with respect to I.
P''(I) = -32

Since the second derivative is negative, the critical point I = 3.75 amperes represents a maximum.

Therefore, the amount of current that will produce the maximum power in the circuit is 3.75 amperes.

To find the maximum power, substitute this value of I back into the original function.
P(I) = 120I - 16I^2
P(3.75) = 120(3.75) - 16(3.75)^2
P(3.75) = 450 - 16(14.06)
P(3.75) = 450 - 224.96
P(3.75) = 225.04 watts

Therefore, the maximum power in the circuit is 225.04 watts.

To find the current that produces the maximum power in the circuit, we need to determine the value of I at which the derivative of the power function P(I) is equal to zero. Let's start by finding the derivative and setting it equal to zero.

P(I) = 120I - 16I^2

Taking the derivative of P(I) with respect to I:

P'(I) = 120 - 2 * 16I

Now, set P'(I) equal to zero and solve for I:

120 - 2 * 16I = 0

Divide both sides of the equation by 2 * 16:

120 = 32I

Divide both sides of the equation by 32:

3.75 = I

Therefore, the current that produces the maximum power in the circuit is approximately 3.75 amperes.

To find the maximum power, we substitute this value of I back into the power function P(I):

P(3.75) = 120(3.75) - 16(3.75)^2

P(3.75) = 450 - 225

P(3.75) = 225

So, the maximum power in the circuit is 225 watts.