Question

The maximum value of the function $f(x)=x\sqrt{4x-x^{2}}$ is ________.

• A. $\sqrt{3}$
• B. $4\sqrt{3}$
• C. $5\sqrt{3}$
• D. $3\sqrt{3}$
• E. $6\sqrt{3}$

Hint:

Check values at critical points and boundaries for absolute maxima.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
To find the maximum value of a function, we use the first derivative test. We find the critical points by setting the first derivative equal to zero. Then we evaluate the function at these critical points and at the endpoints of the domain to find the maximum value.
Step 2: Key Formula or Approach
1. Determine the domain of the function. The expression inside the square root must be non-negative.
2. Find the derivative \(f'(x)\) using the product rule and chain rule.
3. Find the critical points by solving \(f'(x) = 0\).
4. Evaluate the function \(f(x)\) at the critical points and the endpoints of the domain to find the maximum value.
Step 3: Detailed Explanation
1. Determine the domain.
The function \(f(x) = x\sqrt{4x - x^2}\) is defined when \(4x - x^2 \ge 0\).
\(x(4-x) \ge 0\).
The roots are \(x=0\) and \(x=4\). Since the quadratic is downward-opening, the expression is non-negative between the roots.
The domain is \([0, 4]\).
2. Find the derivative \(f'(x)\).
Using the product rule \((uv)' = u'v + uv'\) with \(u=x\) and \(v=\sqrt{4x - x^2}\).
\(u' = 1\).
\(v' = \frac{d}{dx}(4x-x^2)^{1/2} = \frac{1}{2}(4x-x^2)^{-1/2} \cdot (4-2x) = \frac{4-2x}{2\sqrt{4x-x^2}} = \frac{2-x}{\sqrt{4x-x^2}}\).
\[ f'(x) = (1)\sqrt{4x - x^2} + x \left( \frac{2-x}{\sqrt{4x-x^2}} \right) \] Combine the terms by finding a common denominator:
\[ f'(x) = \frac{(\sqrt{4x - x^2})^2 + x(2-x)}{\sqrt{4x-x^2}} = \frac{(4x-x^2) + (2x-x^2)}{\sqrt{4x-x^2}} \] \[ f'(x) = \frac{6x - 2x^2}{\sqrt{4x-x^2}} = \frac{2x(3-x)}{\sqrt{4x-x^2}} \] 3. Find the critical points.
Set \(f'(x) = 0\). This happens when the numerator is zero.
\(2x(3-x) = 0\).
The critical points are \(x=0\) and \(x=3\). Both are within the domain \([0, 4]\).
4. Evaluate the function at critical points and endpoints.
The points to check are the endpoints \(x=0, x=4\) and the critical point \(x=3\).
- \(f(0) = 0\sqrt{4(0) - 0^2} = 0\).
- \(f(4) = 4\sqrt{4(4) - 4^2} = 4\sqrt{16-16} = 0\).
- \(f(3) = 3\sqrt{4(3) - 3^2} = 3\sqrt{12-9} = 3\sqrt{3}\).
Comparing the values {0, 0, \(3\sqrt{3}\)}, the maximum value is \(3\sqrt{3}\).
Step 4: Final Answer
The maximum value of the function is \(3\sqrt{3}\).