maximum value of `log_5(3x+4y)` , if `x^2+y^2=25` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the maximum value of \( \log_5(3x + 4y) \) given the constraint \( x^2 + y^2 = 25 \), we can follow these steps:

### Step 1: Express \( y \) in terms of \( x \)
From the constraint \( x^2 + y^2 = 25 \), we can express \( y \) as:
\[ y = \sqrt{25 - x^2} \]

### Step 2: Substitute \( y \) into the expression
Now, substitute \( y \) into the expression \( 3x + 4y \):
\[ 3x + 4y = 3x + 4\sqrt{25 - x^2} \]

### Step 3: Define a new function
Let \( z = 3x + 4\sqrt{25 - x^2} \). We want to maximize \( z \).

### Step 4: Differentiate \( z \)
To find the maximum value, we differentiate \( z \) with respect to \( x \):
\[ \frac{dz}{dx} = 3 + 4 \cdot \frac{1}{2\sqrt{25 - x^2}} \cdot (-2x) = 3 - \frac{4x}{\sqrt{25 - x^2}} \]

### Step 5: Set the derivative to zero
Set the derivative equal to zero to find critical points:
\[ 3 - \frac{4x}{\sqrt{25 - x^2}} = 0 \]
This simplifies to:
\[ 3 = \frac{4x}{\sqrt{25 - x^2}} \]

### Step 6: Solve for \( x \)
Cross-multiplying gives:
\[ 3\sqrt{25 - x^2} = 4x \]
Squaring both sides:
\[ 9(25 - x^2) = 16x^2 \]
Expanding and rearranging:
\[ 225 - 9x^2 = 16x^2 \implies 225 = 25x^2 \implies x^2 = 9 \]
Thus, \( x = 3 \) or \( x = -3 \). We will take \( x = 3 \) since we want the maximum value.

### Step 7: Find \( y \)
Substituting \( x = 3 \) back into the constraint to find \( y \):
\[ y = \sqrt{25 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \]

### Step 8: Substitute \( x \) and \( y \) back into the expression
Now substitute \( x = 3 \) and \( y = 4 \) into \( 3x + 4y \):
\[ 3(3) + 4(4) = 9 + 16 = 25 \]

### Step 9: Find the maximum value of \( \log_5(3x + 4y) \)
Now we can find the maximum value of the logarithmic expression:
\[ \log_5(25) = \log_5(5^2) = 2 \]

### Conclusion
Thus, the maximum value of \( \log_5(3x + 4y) \) is:
\[ \boxed{2} \]