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Or, Note that in $2 < x < 5$ , $5-x>0, x-2>0$ . Consider the expression $\frac{5-x}{2}+ \frac{5-x}{2}+\frac{x-2}{5} +\frac{x-2}{5} +\frac{x-2}{5} +\frac{x-2}{5} +\frac{x-2}{5} = 3$ And apply AM-GM inequality to these 7 terms.

Equality when , $\frac{x-2}{5} = \frac{5-x}{2} \implies x= \frac{29}{7}$

$Since\quad we\quad need\quad to\quad find\quad the\quad maximum\quad of\quad this\quad equation,\\ so\quad at\quad maxima,\\ derivative\quad of\quad { (5-x) }^{ 2 }{ (x-2) }^{ 5 }\quad w.r.t.\quad x\quad =0\\ So\quad \frac { d }{ dx } { (5-x) }^{ 2 }{ (x-2) }^{ 5 }=0\\ { [(5-x) }^{ 2 }\frac { d }{ dx } { (x-2) }^{ 5 }]+[{ (x-2) }^{ 5 }\frac { d }{ dx } { (5-x) }^{ 2 }=0\\ [5{ (5-x) }^{ 2 }{ (x-2) }^{ 4 }\frac { d }{ dx } (x-2)]+[{ 2(x-2) }^{ 5 }(5-x)\frac { d }{ dx } (5-x)]=0\\ [5{ (5-x) }^{ 2 }{ (x-2) }^{ 4 }(1)]+[2{ (x-2) }^{ 5 }(5-x)(-1)]=0\\ Taking\quad common:\\ { (x-2) }^{ 4 }(5-x)\quad [5(5-x)-2(x-2)]=0\\ { (x-2) }^{ 4 }(5-x)(25-5x-2x+4)=0\\ { (x-2) }^{ 4 }(5-x)(29-7x)=0\\ Clearly,\quad either\quad x=2,\quad x=5\quad or\quad x=\frac { 29 }{ 7 } \\ Now,\quad it\quad has\quad been\quad given\quad that\quad 2\quad <\quad x\quad <\quad 5,so,\quad x\quad is\quad \\ only\quad left\quad with\frac { 29 }{ 7 } \\ Comparing\quad with\quad x=\frac { a }{ b } ,\quad a=29,\quad b=7.\\ So\quad a\quad \times \quad b=\quad 29\times 7\quad =\quad 203$

CHEERS!!:):)

One comment. You should probably mention that a and b are positive coprime integers.

$(5-x)^2(x-2)$

Substitute $u=x-2 <=> -x=3-u$ and the equation reduces tremendously:

$u^5(3-u)^2$

$u^5(9-6u+u^2)$

$f(u)=u^7-6u^6+9u^5$

This leads to:

$f'(u)=7u^6(u^2-\frac{36}{7}u+\frac{45}{7})$

$u$ can't be $0$ because of the restrictions on the interval. This results to a simple quadratic equation:

$u^2-\frac{36}{7}u+\frac{45}{7}$

We get

$u=\frac{15}{7} \wedge u=3$

Coming back to our substitution we recognize that $u=3$ is not in the given intervall, thus:

$\frac{15}{7}=x-2 <=> \boxed{x=\frac{29}{7}}$

Use the concept of maxima and minima to find the answer

Let the function be $f(x)$ , so that $f(x)=(5-x)^2(x-2)^5$ .

Differentiating using the product rule and chain rule gives:

$f'(x)=5(x-2)^4(5-x)^2-2(5-x)(x-2)^5$

Setting $f'(x)=0$ to find maximum value:

$5(x-2)^4(5-x)^2-2(5-x)(x-2)^5=0$

$5(x-2)^4(5-x)^2=2(5-x)(x-2)^5$

Simplifying:

$5(5-x)=2(x-2)$

$25-5x=2x-4$

$7x=29$

$x=29/7$

Comparing with $x=a/b: a=29 ; b=7$

Therefore, $ab=29×7=203$

On a side note, you should state that a and b are positive coprime integers, or else they can be any two numbers which when a divided by b gives 29/7.

Differentiate as follows:

$\dfrac{d}{dx}\left((5-x)^2(x-2)^5\right)=$

$-2(5-x)(x-2)^5 + (5-x)^2 5(x-2)^4 =$

$(5-x)(x-2)^4(7x-29)$

so solving for $x$ in

$7x-29=0$

gets us $x=\dfrac{29}{7}$

Similar solution with @Shivang Jindal 's but different presentation.

Since $5-x > 0$ and $x-2 > 0$ for $2< x < 5$ , we can apply AM-GM inequality as follows:

$\begin{aligned} \frac {5-x}2 + \frac {5-x}2 + \frac {x-2}5 + \frac {x-2}5 + \frac {x-2}5 + \frac {x-2}5 + \frac {x-2}5 + \frac {x-2}5 & \ge 7 \sqrt[7]{\left(\frac {5-x}2 \right)^2 \left( \frac {x-2}5\right)^5} \end{aligned}$

Equality occurs when $\dfrac {5-x}2 = \dfrac {x-2}5$ or $x = \dfrac {29}7$ . Therefore $ab = 29 \times 7 = \boxed{203}$ .