Quadratic functions, when finding maximum and minimum value.?

use differential calculus

P' = 120 - 32i

at p max, the rate of change is equal to 0.

therefore, at p max, P' = 0

when 120 - 32i = 0

32i = 120

i = 15/4

plug the value of i back to the original function:

when i = 15/4

p = 120 (15/4) - 16 (15/4)^2

p = 450 - 225

p = 225

I have forgotten my physics but this looks like pure calculus.

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

P '(i) = 120 - 32i

Set this equal to 0.

120 - 32i = 0

120 = 32i

32i = 120

i = 120/32 = [(8*15) / (8*4)] = 15/4.....a local extremum (that is a maximum, or a minimum).

Take the second derivative of P(i)

P"(i) = -32. Since this is negative the point where x = 15/4 is a local maximum, and the only maximum.

P(15/4) = (120 * (15/4)) - (16 * (15/4) * (15/4))

120 * (15/4) = 30 * 15 = 450

16* (15/4) * (15/4) = 15 * 15 = 225

450 - 225 = 225

So the maximum occurs at (15/4, 225)

The maximum power is 225 watts.

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If you prefer to solve this with analytic geometry you could do the following:

If you think of P as y and i as x, then this equation is a parabola that opens down. The vertex which is (15/4, 225) is again the maximal point and the answer comes out to be 225 watts.

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Take your pick of methods :D - I prefer using calculus.

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I uploaded that "Power Parabola" to photobucket. If you want to see it click on:

http://i369.photobucket.com/albums/oo133/gerryrain...

Notice that the x-axis and y-axis have different measures. Otherwise x = 15/4 = 3.75 and y = 225 would be totally unrecognizable.

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.

No calculus needed in this problem.

The vertex of any parabola occurs when the independent variable (x, or i, in this case) is equal to -b/(2a), when the quadratic is written in standard form.

In this problem, b = 120 and a = -16.

i = -b/(2a) = -120/(2*-16) = 120/32 = 15/4 amperes

The maximum power will be:

P = 120(15/4) - 16(15/4)^2 = 225 watts

The vertex form of a vertical parabola's equation is generally expressed as: y = a(x - h)² + k where (h,k) is the vertex.

p = -16i² + 120i ........ rewrite in vertex form

= -16(i² - 15/2i)

= -16((i - 15/4)² - 225/16)

= -16(i - 15/4)² + 225

The vertex of this parabola is at (15/4,225)

Or you can use calculus:

p = -16i² + 120i

dp/di = -32i + 120

d²p/di² = -32

32i =120 ........ set dp/di = 0 to find stationary points

i = 15/4 ........... d²p/di² < 0, ⇒15/4 is a relative max.

Answer: max power is 225 W at 15/4 Amps.