The minimum value of `x^(2) + (250)/(x)` is

Options

A. 75
B. 55
C. 50
D. 25

Answer

To find the minimum value of the function \( f(x) = x^2 + \frac{250}{x} \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}\left(\frac{250}{x}\right) \]

Using the power rule and the quotient rule, we get:

\[ f'(x) = 2x - \frac{250}{x^2} \]

### Step 2: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:

\[ 2x - \frac{250}{x^2} = 0 \]

### Step 3: Solve for \( x \)
Rearranging the equation gives:

\[ 2x = \frac{250}{x^2} \]

Multiplying both sides by \( x^2 \):

\[ 2x^3 = 250 \]

Dividing both sides by 2:

\[ x^3 = 125 \]

Taking the cube root of both sides:

\[ x = 5 \]

### Step 4: Determine if it is a minimum or maximum
Next, we need to check whether this critical point is a minimum or maximum by using the second derivative test.

First, we find the second derivative \( f''(x) \):

\[ f''(x) = \frac{d}{dx}(f'(x)) = \frac{d}{dx}\left(2x - \frac{250}{x^2}\right) \]

Calculating the second derivative:

\[ f''(x) = 2 + \frac{500}{x^3} \]

Now, we evaluate \( f''(5) \):

\[ f''(5) = 2 + \frac{500}{5^3} = 2 + \frac{500}{125} = 2 + 4 = 6 \]

Since \( f''(5) > 0 \), this indicates that \( x = 5 \) is a local minimum.

### Step 5: Calculate the minimum value
Now, we find the minimum value of the function by substituting \( x = 5 \) back into the original function:

\[ f(5) = 5^2 + \frac{250}{5} = 25 + 50 = 75 \]

Thus, the minimum value of \( f(x) \) is:

\[ \boxed{75} \]