2nd derivative test , Maximum point A study shows that a human body reaction R to a dosage D of a certain drug is given by 
$$ R = D^2(\frac{k}{2} - \frac{D}{3}) $$ 
Show that he rate of change in the reaction $R$ with respect to the dosage $D$ is maximum if $D = K/2 $ 
My first derivative is 
$R' = KD - D^2 $,
Dosage for maximum reaction = $k$
My second derivative is 
$R''= K - 2D $
$R(\frac{k}{2}) = k - 2(\frac{k}{2}) = 0 $ 
But 0 is neither negative or positive , how do I show that it's a maximum (negative) ? 
 A: Use the first derivative test on $R'$. It's immediate that $R''>0$ when $D < k/2$, and $R''<0$ when $D>k/2$. Thus, since $R''(k/2)=0$, it follows that $R'$ has a local maximum at $D=k/2$. But since there's only one critical point, it follows that the local maximum is, in fact, a global maximum.

However, if you prefer, you can use the second derivative test. The second derivative of $R'$ is $R'''$ which is identically $-2$. In particular, $R'''(k/2)<0$, hence, since $R''(k/2)=0$, it follows that $R'$ has a local maximum at $D=k/2$. But once again, since there's only one critical point, it follows that the local maximum is, in fact, a global maximum.

An even simpler approach is to note that $R'$ is a quadratic function of $D$ with negative leading coefficient, hence the critical value for $R'$ (i.e., $D=k/2)$ is where $R'$ attains a global maximum.
A: As I mentioned in my comment, you are confused with what the question being asked is. It wants you to show that the rate of change, $R'$, (not $R$) achieves a maximum at $\frac{k}{2}$.
By setting $R''=0$, you have shown that there is a critical point at $\frac{k}{2}$, now you only have to show that it is a maximum.
I will take the approach of the second derivative test as you mentioned in the title.
We must find the second derivative of $R'$, namely, $R'''$.
$R'''=\frac{d}{dD}(R'')=\frac{d}{dD}(K-2D)=-2.$
Since $R'''$, the second derivative of $R'$, is negative regardless of what $D$ is, then $D= \frac{k}{2}$ which we found as the only critical point is a maximum.
A: D = k or 0 (don't miss the 0 solution!) satisfy the first-order condition; beware that this doesn't guarantee max/min yet (i.e. FOC is just a necessary condition).  So don't jump to any conclusion at this stage.
k/2 doesn't satisfy FOC, so it definitely isn't a max/min.
FYI, if some day you encounter a 0 when checking the second order condition, it means the FOC solutions may be max/min/a saddle point.  Essentially, you can't infer anything from the second-order test.
