Question

The minimum value of $\left( {{x^2} + \dfrac{{250}}{x}} \right)$ is:
A. $75$
B. $50$
C. $25$
D. $55$

Answer 1

These type questions can be easily solved if we keep in mind the concepts of maxima and minima. In the problem, we are required to find the minimum value of the function in the variable x, $\left( {{x^2} + \dfrac{{250}}{x}} \right)$. We do so by differentiating the function with respect to x and finding the critical points at which the function attains maximum or minimum values. Then, we use the second derivative test to check whether the critical point represents the minimum or maximum value of the function.

Complete step by step answer:
So, we have the function in x as $\left( {{x^2} + \dfrac{{250}}{x}} \right)$. Let us assume the function $\left( {{x^2} + \dfrac{{250}}{x}} \right)$ to be equal to y. So,
$y = \left( {{x^2} + \dfrac{{250}}{x}} \right)$
Now, we differentiate both sides of the above equation. So, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^2} + \dfrac{{250}}{x}} \right)$
We know the power rule of differentiation $\dfrac{{d\left[ {{x^n}} \right]}}{{dx}} = n{x^{\left( {n - 1} \right)}}$. So, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{d}{{dx}}\left( {\dfrac{{250}}{x}} \right)$
$\Rightarrow \dfrac{{dy}}{{dx}} = 2x - \dfrac{{250}}{{{x^2}}} - - - - \left( 1 \right)$

Now, we equate the first derivative of the function to zero to find the critical points of the function. So, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = 2x - \dfrac{{250}}{{{x^2}}} = 0$
Dividing both sides of the equation by $2$, we get,
$\Rightarrow x - \dfrac{{125}}{{{x^2}}} = 0$
Taking LCM of the denominators, we get,
$\Rightarrow \dfrac{{{x^3} - 125}}{{{x^2}}} = 0$
Multiplying both sides of equation by ${x^2}$, we get,
$\Rightarrow {x^3} - 125 = 0$
Taking co0nstant term to right side of equation,
$\Rightarrow {x^3} = 125 = {5^3}$
Taking cube root on both sides of equation, we get,
$\Rightarrow x = 5$
So, we get $x = 5$ as a critical point of the function.

Now, we find the second derivative of the function. From equation $\left( 1 \right)$, we have,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{dy}}{{dx}}\left[ {2x - \dfrac{{250}}{{{x^2}}}} \right]$
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{dy}}{{dx}}\left[ {2x - 250{x^{ - 2}}} \right]$
Again, using the power rule of differentiation, we get,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2 + 500{x^{ - 3}}$
Now, substituting the value of x as $5$,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2 + 500{\left( 5 \right)^{ - 3}}$
Now, we know that ${5^{ - 3}} = \dfrac{1}{{125}}$. So, we get,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2 + \dfrac{{500}}{{125}}$
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2 + 4 = 6$
So, the second derivative of the function is positive for $x = 5$. So, $x = 5$ is the point of local minima. So, the minimum value of function is attained at $x = 5$. So, substituting value of x as $5$ in the original function, we get,
$\Rightarrow \left( {{5^2} + \dfrac{{250}}{5}} \right)$
$\Rightarrow \left( {25 + 50} \right)$
$\Rightarrow 75$
So, the minimum value of the function $\left( {{x^2} + \dfrac{{250}}{x}} \right)$ is $75$.

Therefore, option A is the correct answer.

Note: We can solve the problems involving the maxima and minima concept by two methods: the first derivative test and the second derivative test. The first derivative test helps in finding the local extremum points of the function. The second derivative test involves finding the local extremum points and then finding out which of them is local minima and which is local maxima. Take care while handling calculative steps so as to be sure of the final answer.