Explanation

(c)
$p(t) = 1000 + \frac{1000t}{100 + t^{2}}$
$\Rightarrow \ \ \ \ \ \ \ p^{'}(t) = 0 + \frac{\left( 100 + t^{2} \right)(1000) - 1000t\ (2t)}{\left( 100 + t^{2} \right)^{2}}$
$= 1000\frac{(100 - t^{2})}{\left( 100 + t^{2} \right)^{2}}$
Put $p^{'}(t) = 0$ for maxima or minima
$\Rightarrow \ \ \ \ \ \ 100 - t^{2} = 0$
$\Rightarrow \ \ \ \ \ t = \pm 10$
Now, $p^{''}(t) = 1000$
$\times \left\lbrack \frac{\left( 100 + t^{2} \right)^{2}( - 2t) - \left( 100 - t^{2} \right)2(100 + t^{2})2t}{\left( 100 + t^{2} \right)^{4}} \right\rbrack$
$= 1000t\frac{\lbrack\left( 100 + t^{2} \right)( - 2) - (100 - t^{2})(4)\rbrack}{\left( 100 + t^{2} \right)^{3}}$
$= - 1000t\ \frac{\lbrack 600 - 2t^{2}\rbrack}{\left( 100 + t^{2} \right)^{3}}$
At $t = 10,\ \ \ p^{''}(t) < 0$
$\therefore$ The maximum value is
$p(10) = 1000 + \frac{10000}{100 + 100}$
$= 1000 + \frac{10000}{200} = 1050$