1. ## Finding maximum

$y=3x-\frac{x^2}{2}$

Find y max

Here is how I'm approaching this ...

$6x-x^2$

$y'=6-2x$

With 3 being a critical point

$y''=-2$

So the critical point is a maximum. It's a parabola opening downward from 3?

2. We have $y = 3x - \frac{x^2}{2}$

But
$6x-x^2$ ??
$y' = 6-2x$ ??
These are wrong.

But I see what you were trying to do.

It should be,

$y = 3x - \frac{x^2}{2}$

$2y = 6x - x^2$

$2y' = 6 - 2x$

$y' = 3 - x$

I know, the root doesn't change by multiplication or division, but $y' = 6-2x$ is not correct.

3. Thank you for clarifying. I'm prone to making mistakes and that is why I like to double check. Looking at the equation is still seems 3 is our critical point, but the differences in our y' will become ...

$y''=-1$

Which should still give us 3 as our maximum with a downward opening parabola. I appreciate the help as this mistake could have cost me in other equations

4. Are you just trying to find the max of $3x-\frac{x^{2}}{2}$?.

Did you just differentiate?.

$y'=3-x$

3-x=0, x=3

That gives y=9/2 as the max value for y.

If I am misunderstanding, I apologize.

5. Originally Posted by XIII13Thirteen
Which should still give us 3 as our maximum with a downward opening parabola. I appreciate the help as this mistake could have cost me in other equations
Do you know that the maximum is at $x=3$ and the maximum value of the function is $f(3) = 9/2$ as galactus said? If you know it, no problem

6. You can also find this without calc by using $x=\frac{-b}{2a}$.

Which gives the x-coordinate for the vertex of a parabola.

In your case, a=-1/2 and b=3

$\frac{-3}{2(\frac{-1}{2})}=3$

The equation wanted "y max," not "x max" so technically it is $\frac{9}{2}$ is correct. I'm doing these equations with a multiple choice solution set and he has

a.)4

b.)3

c.)-1

d.)none of these

I originally picked b because I was thinking of in terms of x, but in the context of "find y max" $\frac{9}{2}$. Evil questions. Thanks for catching that