Find the two positive numbers x and y whose sum is 35 and the product `x^(2) y^(5)` maximum.

Answer

To find the two positive numbers \( x \) and \( y \) whose sum is 35 and the product \( x^2 y^5 \) is maximized, we can follow these steps:

### Step 1: Set up the equations
We know from the problem that:
\[ x + y = 35 \]
We want to maximize the product:
\[ P = x^2 y^5 \]

### Step 2: Express one variable in terms of the other
From the first equation, we can express \( y \) in terms of \( x \):
\[ y = 35 - x \]

### Step 3: Substitute \( y \) in the product equation
Now, substitute \( y \) into the product equation:
\[ P = x^2 (35 - x)^5 \]

### Step 4: Differentiate the product
To find the maximum, we need to differentiate \( P \) with respect to \( x \):
\[ P' = \frac{d}{dx}(x^2 (35 - x)^5) \]
Using the product rule:
\[ P' = 2x(35 - x)^5 + x^2 \cdot 5(35 - x)^4 \cdot (-1) \]
Simplifying this gives:
\[ P' = 2x(35 - x)^5 - 5x^2(35 - x)^4 \]

### Step 5: Factor the derivative
We can factor out common terms:
\[ P' = (35 - x)^4 \left[ 2x(35 - x) - 5x^2 \right] \]
This simplifies to:
\[ P' = (35 - x)^4 (70x - 2x^2 - 5x^2) = (35 - x)^4 (70x - 7x^2) \]

### Step 6: Set the derivative to zero
To find critical points, set \( P' = 0 \):
\[ (35 - x)^4 (70x - 7x^2) = 0 \]
This gives us two cases:
1. \( 35 - x = 0 \) which gives \( x = 35 \) (not valid since \( y \) must be positive)
2. \( 70x - 7x^2 = 0 \)

Factoring out \( 7x \):
\[ 7x(10 - x) = 0 \]
This gives us \( x = 0 \) or \( x = 10 \).

### Step 7: Find corresponding \( y \)
If \( x = 10 \):
\[ y = 35 - 10 = 25 \]

### Step 8: Verify maximum
To confirm that this gives a maximum, we can check the second derivative or analyze the sign of \( P' \) around \( x = 10 \).

### Conclusion
The two positive numbers that maximize the product \( x^2 y^5 \) under the constraint \( x + y = 35 \) are:
\[ x = 10, \quad y = 25 \]