<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] $P(x)=x+a(x-2)(x-4)(x-6)$ because $2,4,6$ are roots of the cubic $P(x)-x$, whose leading coefficient is $a$.
Therefore, $P(x)=a x^3 - 12 a x^2 + (44 a + 1) x - 48 a$.
Since the coefficients are all at most $12$, we need
$$
a \le 12, \quad -12 a \le 12, \quad 44a+1 \le 12, \quad -48a \le 12
$$
Therefore,
$
-\frac14 \le a \le \frac14
$.
Now, $P(10)=10+192a$, whose maximum value is $58$, attained at $a=\frac14$.
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