Answer:
The current that produces maximum power is 3A
Step-by-step explanation:
Given
[tex]P(c) = 20(c - 3)^2[/tex]
Required [Missing from the question]
The current that produces maximum power
First, we represent the function in standard form
[tex]P(c) = 20(c - 3)^2[/tex]
[tex]P(c) = 20(c - 3)(c - 3)[/tex]
Open bracket
[tex]P(c) = 20(c^2 -6c+ 9)[/tex]
[tex]P(c) = 20c^2 -120c+ 180[/tex]
The maximum value of c is:
[tex]Max(c) = \frac{-b}{2a}[/tex]
Where:
[tex]f(x) = ax^2 + b^2 + c[/tex]
By comparison: [tex]P(c) = 20c^2 -120c+ 180[/tex]
[tex]a = 20[/tex]
[tex]b = -120[/tex]
[tex]c = 180[/tex]
So, we have:
[tex]Max(c) = \frac{-b}{2a}[/tex]
[tex]Max(c) = \frac{-(-120)}{2 * 20}[/tex]
[tex]Max(c) = \frac{120}{40}[/tex]
[tex]Max(c) = 3[/tex]
