There are $20$ books on Algebra & Calculus in our library. Prove that the greatest number of selections each of which consists of $5$ books on each topic is possible only when there are $10$ books on each topic in the library.
My attempt is as follows:-
Let there are $x$ books of Algebra and $20-x$ books of Calculus
So number of selections each of which consists of $5$ books on each topic= $\displaystyle{x\choose 5}{20-x\choose 5}$
$$y=\dfrac{x(x-1)(x-2)(x-3)(x-4)(20-x)(19-x)(18-x)(17-x)(16-x)}{5!\cdot5!}$$
Let's take pairs $x(20-x),(x-1)(19-x),(x-2)(18-x),(x-3)(17-x),(x-4)(16-x)$
Let's find the maximum value of $(x-1)(19-x)$
$$y=19x-x^2-19+x$$ $$y=-x^2+20x-19$$ $$\dfrac{dy}{dx}=-2x+20$$
$$2x=20,x=10$$
In the same way all the other pairs will get maximum value at $x=10$
So can we say that maximum value of the expression will be at $x=10$? Is there any better approach than this?
Since $x$ will go from 5 to 15, ${x \choose 5}$ will increase while ${20-x \choose 5}$ will decrease as $x$ increases. You can intuitively arrive at the conclusion that the product will be maximum at $x=10$.