Question

The minimum value of the function $f(x) = x^2 + \frac{250}{x}$ for $x > 0$ is:

• A. $75$
• B. $50$
• C. $25$
• D. $100$

Hint:

Alternatively, you can use the AM-GM inequality: $x^2 + \frac{125}{x} + \frac{125}{x} \ge 3\sqrt[3]{x^2 \cdot \frac{125}{x} \cdot \frac{125}{x}} = 3(25) = 75$. This avoids calculus entirely!

Answer 1

The Correct Option is A

Step 1: Plan how to find the minimum.
To find the smallest value of a smooth function, we look for points where its slope is zero, then check that it really is a low point. So we differentiate, set it to zero, and confirm.

Step 2: Differentiate the function.
The function is $f(x) = x^2 + \frac{250}{x}$. Write $\frac{250}{x}$ as $250x^{-1}$ before differentiating.
\[ f'(x) = 2x - \frac{250}{x^2} \]

Step 3: Set the slope to zero.
A flat point has zero slope.
\[ 2x - \frac{250}{x^2} = 0 \implies 2x = \frac{250}{x^2} \]

Step 4: Solve for $x$.
Cross multiply to get $2x^3 = 250$, so $x^3 = 125$, giving $x = 5$. Since $x > 0$, this is allowed.

Step 5: Confirm it is a minimum.
The second derivative is $f''(x) = 2 + \frac{500}{x^3}$. At $x = 5$ this is $2 + \frac{500}{125} = 6$, which is positive, so the point is indeed a minimum.

Step 6: Find the minimum value.
Put $x = 5$ back into the original function.
\[ f(5) = 25 + \frac{250}{5} = 25 + 50 = 75 \]
\[ \boxed{75} \]