Question

The power generated by an electrical circuit (in watts) as a function of its current ccc (in amperes) is modeled by: P(c)=-20(c-3)^2+180P(c)=−20(c−3) 2 +180

Answer 1

Answer:

The current that produces maximum power is 3A

Step-by-step explanation:

Given

$P(c) = 20(c - 3)^2$

Required [Missing from the question]

The current that produces maximum power

First, we represent the function in standard form

$P(c) = 20(c - 3)^2$

$P(c) = 20(c - 3)(c - 3)$

Open bracket

$P(c) = 20(c^2 -6c+ 9)$

$P(c) = 20c^2 -120c+ 180$

The maximum value of c is:

$Max(c) = \frac{-b}{2a}$

Where:

$f(x) = ax^2 + b^2 + c$

By comparison: $P(c) = 20c^2 -120c+ 180$

$a = 20$

$b = -120$

$c = 180$

So, we have:

$Max(c) = \frac{-b}{2a}$

$Max(c) = \frac{-(-120)}{2 * 20}$

$Max(c) = \frac{120}{40}$

$Max(c) = 3$