<s>[INST] Maximum possible value of $P(10)$ The real numbers $a$, $b$, $c$, and $d$ are each less than or equal to $12$.  The polynomial $$P(x)=ax^3+bx^2+cx+d$$ satisfies $P(2)=2$, $P(4)=4$ and $P(6)=6$. Find the maximum possible value of $P(10)$.
What I did was first I used the given information to get $3$ equation in $a, b,c,d$. Then I obtained values of $a$, $b$, $c$ in terms of $d$.  Then using these values, I found $P(10)$ in terms of $d$ and substituted $d=12$. But I couldn't arrive at the answer.  
Thanks in advance 
 [/INST] From the system
$
\left\{
\begin{array}{l}
 8 a+4 b+2 c+d=2 \\
 64 a+16 b+4 c+d=4 \\
 216 a+36 b+6 c+d=6 \\
\end{array}
\right.
$
we get 
$b = -12 a, c = 1 + 44 a, d = -48 a$
with the constraints
$-12 a\leq 12\land 44 a+1\leq 12\land -48 a\leq 12$
which give the limitation
$-\frac{1}{4}\leq a\leq \frac{1}{4}$
Therefore plugging in the polynomial we have
$P(x)=a x^3-12 a x^2+(44 a+1) x-48 a$
and $P(10)=2 (96 a+5)$
which is maximum for $a=\frac{1}{4}$
and $P(10)=58$
Hope this is useful
</s>