Question

If \( x+y=50 \), then the maximum value of \( \sqrt{4xy} \) is

• A. \( 25 \)
• B. \( 50 \)
• C. \( 100 \)
• D. \( 625 \)
• E. \( 2500 \)

Hint:

Whenever the sum of two numbers is fixed, their product is maximum when the two numbers are equal. This is a direct application of AM-GM inequality.

Answer 1

The Correct Option is B

Step 1: Write the given condition clearly.
We are given \[ x+y=50 \] and we need to find the maximum value of \[ \sqrt{4xy} \] Since the square root function is increasing for non-negative values, maximizing \( \sqrt{4xy} \) is equivalent to maximizing \( 4xy \), or simply maximizing \( xy \).

Step 2: Express the target in a simpler form.
Observe that \[ \sqrt{4xy}=2\sqrt{xy} \] So the problem reduces to finding the maximum value of \( xy \) under the condition \( x+y=50 \).

Step 3: Use the standard result for fixed sum.
For two real numbers with fixed sum, the product is maximum when the two numbers are equal.
Thus, under \[ x+y=50 \] the product \( xy \) is maximum when \[ x=y=25 \]

Step 4: Verify this using AM-GM inequality.
By AM-GM, \[ \frac{x+y}{2}\geq \sqrt{xy} \] Substituting \( x+y=50 \), we get \[ \frac{50}{2}\geq \sqrt{xy} \] \[ 25\geq \sqrt{xy} \] Squaring both sides, \[ 625\geq xy \] So the maximum possible value of \( xy \) is \[ 625 \] and this occurs when \[ x=y=25 \]

Step 5: Find the value of \( \sqrt{4xy} \).
Now substitute \( xy=625 \): \[ \sqrt{4xy}=\sqrt{4\cdot 625} \] \[ =\sqrt{2500} \] \[ =50 \]

Step 6: Check with \( x=y=25 \).
If \[ x=25,\qquad y=25 \] then \[ x+y=50 \] and \[ \sqrt{4xy}=\sqrt{4\cdot 25\cdot 25}=\sqrt{2500}=50 \] So the value is valid and attainable.

Step 7: Final conclusion.
Hence, the maximum value of \[ \sqrt{4xy} \] is \[ \boxed{50} \] Therefore, the correct option is \[ \boxed{(2)\ 50} \]