Find `max^(m)&min^(m)` value of `27^(sinx).81^(cosx)`

Answer

To find the maximum and minimum values of the expression \( 27^{\sin x} \cdot 81^{\cos x} \), we can follow these steps:

### Step 1: Rewrite the expression in terms of base 3
We start by rewriting \( 27 \) and \( 81 \) in terms of base \( 3 \):
\[ 27 = 3^3 \quad \text{and} \quad 81 = 3^4 \]
Thus, we can express the original expression as:
\[ 27^{\sin x} \cdot 81^{\cos x} = (3^3)^{\sin x} \cdot (3^4)^{\cos x} \]

### Step 2: Apply the power of a power property
Using the property of exponents \( (a^m)^n = a^{m \cdot n} \), we can simplify further:
\[ (3^3)^{\sin x} = 3^{3 \sin x} \quad \text{and} \quad (3^4)^{\cos x} = 3^{4 \cos x} \]
Therefore, we can combine these:
\[ 27^{\sin x} \cdot 81^{\cos x} = 3^{3 \sin x} \cdot 3^{4 \cos x} = 3^{3 \sin x + 4 \cos x} \]

### Step 3: Find the maximum and minimum of the exponent
Now, we need to find the maximum and minimum values of the expression \( 3 \sin x + 4 \cos x \). This expression can be analyzed using the formula for the maximum and minimum values of \( a \sin x + b \cos x \), which is given by:
\[ \text{Max} = \sqrt{a^2 + b^2} \quad \text{and} \quad \text{Min} = -\sqrt{a^2 + b^2} \]
Here, \( a = 3 \) and \( b = 4 \).

### Step 4: Calculate \( \sqrt{a^2 + b^2} \)
Calculating \( \sqrt{3^2 + 4^2} \):
\[ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 5: Determine the maximum and minimum values
Thus, the maximum and minimum values of \( 3 \sin x + 4 \cos x \) are:
\[ \text{Max} = 5 \quad \text{and} \quad \text{Min} = -5 \]

### Step 6: Substitute back to find the maximum and minimum values of the original expression
Now we substitute these values back into our expression for \( 27^{\sin x} \cdot 81^{\cos x} \):
- Maximum value:
\[ 3^{5} = 243 \]
- Minimum value:
\[ 3^{-5} = \frac{1}{3^5} = \frac{1}{243} \]

### Final Result
Thus, the maximum value is \( 243 \) and the minimum value is \( \frac{1}{243} \).