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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

Math Lover
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Rohan Shinde
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  • Note that maximizing $d$ will not necessarily maximize $P(10)$ since $a$, $b$, and $c$ all depend on $d$. It's possible that maximizing $d$ will, for example, make $a$ so small that, for example, $P(10)$ would be larger if $d$ were, for example, $10$. For example. –  Nov 15 '17 at 18:50
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    Please show the expression you got for $P(10)$ as a function of $d$ – Ross Millikan Nov 15 '17 at 18:52

3 Answers3

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$P(x)=x+a(x-2)(x-4)(x-6)$ 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 $a=\frac14$.

lhf
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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

Raffaele
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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.

Ross Millikan
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