A missile fired from ground level rises $x$ metres vertically upwards in $t$ seconds and $x=t\left( 100-12.5t \right)$. Then the maximum height reached by the missile is.
(a) $100\text{ m}$
(b) $150\text{ m}$
(c) $250\text{ m}$
(d) $200\text{ m}$

Hint: For solving this question we will use the condition of maxima and minima to find the maximum value of height reached by the missile with the help of the equation given in the question.

Complete step-by-step solution -

Given:
It is given that a missile fired from ground level rises $x$ metres vertically upwards in $t$ seconds and $x=t\left( 100-12.5t \right)$.
Now, we can write that height $x$ can be written in the form of $x=f\left( t \right)=t\left( 100-12.5t \right)$.
Before we proceed we should know how the derivative of the function $f\left( x \right)$ helps us in determining the maximum and minimum value of the function. If the maximum and minimum value of $f\left( x \right)$ occurs at $x={{x}_{1}}$ and $x={{x}_{2}}$ respectively. Then,
1. ${f}'\left( {{x}_{1}} \right)={{\left[ \dfrac{d\left( f\left( x \right) \right)}{dx} \right]}_{x={{x}_{1}}}}=0$ and ${f}''\left( {{x}_{1}} \right)={{\left[ \dfrac{d\left( {f}'\left( x \right) \right)}{dx} \right]}_{x={{x}_{1}}}}<0$.
2. ${f}'\left( {{x}_{2}} \right)={{\left[ \dfrac{d\left( f\left( x \right) \right)}{dx} \right]}_{x={{x}_{2}}}}=0$ and ${f}''\left( {{x}_{2}} \right)={{\left[ \dfrac{d\left( {f}'\left( x \right) \right)}{dx} \right]}_{x={{x}_{2}}}}>0$.
Now, in this question, we have to find the value of $t$ for which the value of $x=f\left( t \right)$ will be maximum. Then,
$\begin{align}
  & f\left( t \right)=t\left( 100-12.5t \right) \\
 & \Rightarrow f\left( t \right)=100t-12.5{{t}^{2}} \\
\end{align}$

$\begin{align}
  & \Rightarrow {f}'\left( t \right)=100-25t=0 \\
 & \Rightarrow 100=25t \\
 & \Rightarrow t=4 \\
\end{align}$
Now, as we have got $t=4$ for which the first derivative of $f\left( t \right)=t\left( 100-12.5t \right)$ is zero. So, we can check whether the second derivative is positive or negative for $t=4$ . Then,
$\begin{align}
  & f\left( t \right)=t\left( 100-12.5t \right) \\
 & \Rightarrow f\left( t \right)=100t-12.5{{t}^{2}} \\
 & \Rightarrow {f}'\left( t \right)=100-25t \\
 & \Rightarrow {f}''\left( t \right)=-25 \\
 & \Rightarrow {f}''\left( t \right)<0 \\
\end{align}$
Now, as the second derivative is negative so, at $t=4$ the function $f\left( t \right)=t\left( 100-12.5t \right)$ will be maximum.
Now, we find the value of $f\left( t \right)=t\left( 100-12.5t \right)$ at $t=4$ . Then,
$\begin{align}
  & f\left( t \right)=t\left( 100-12.5t \right) \\
 & \Rightarrow f\left( 4 \right)=4\left( 100-12.5\times 4 \right) \\
 & \Rightarrow f\left( 4 \right)=4\times 50 \\
 & \Rightarrow f\left( 4 \right)=200 \\
\end{align}$
Now, from the above result, we can say that $x$ will be maximum when $t=4$ .
Thus, the missile will reach maximum up to a height of 200 m.
Hence, (d) is the correct option.

Note: Here, the students should first understand what we have to find then proceed in the right directions. Moreover, we should apply the concept of maxima and minima accurately and don’t confuse the value of the first derivative and second derivative for each case. Then, equations should be solved stepwise to prove the result.