Calculus problem. maximum value of "y"?

Question

if y = 24x + 3x^2 - x^3, what is the relative maximum value of y?

so far I've got:
1) derivative
2) derivative = 0, solve
3) quadratic formula
4) bang head on desk because it still didn't work.

please help =.O

Answer 1

As you know, you need to set the derivative to 0 to solve this.

y = 24x + 3x^2 - x^3

y' = 24 + 6x - 3x^2 = 0

y' = -3(x^2-2x-8) = -3(x-4)(x+2) = 0

x = 4 and x = -2 make it zero

The relative maximum occurs where y'' < 0

y'' = 6 - 6x

y''(4) = 6-6*4 = -18
y''(-2) = 6-6(-2) = 18

The relative maximum occurs where x = 4,
and the relative maximum value
y(4) = 24(4) + 3(4)^2 - (4)^3
= 96 + 48 - 64
= 80

Answer 2

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Answer 3

y´= -3x^2+6x+24
x^2-2x-8=0 x= ((2+-sqrt(36)/2
x=4 and x= -2
sign y´ --------- (-2)++++++++(4)---------
so at x= 4 there is a relative maximum
y(4) =96+48 -64= 80

Answer 4

diffrentiate it