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 
 A: $P(x)=x+a(x-2)(x-4)(x-6)$ 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 $a=\frac14$.
A: 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
A: Once you find $P(10)$ as a function of $d$ you want to find the value of $d$ that maximizes $P(10)$ subject to the restriction $d \le 12$, so take the derivative with respect to $d$, set to zero, solve for $d$.  Then compare with the $P(10)$ that comes from $d=12$ because the optimum may come at the boundary.
