QPaperGen

We are given:
\[
x + y = 35 \Rightarrow y = 35 - x
\]

We want to maximize:
\[
P(x) = x^2 (35 - x)^5
\]

Differentiate using the product rule:
\[
P'(x) = \frac{d}{dx}\left[ x^2 (35 - x)^5 \right] = 2x(35 - x)^5 + x^2 \cdot 5(35 - x)^4 \cdot (-1)
\]
\[
P'(x) = 2x(35 - x)^5 - 5x^2(35 - x)^4
\]
Factor common terms:
\[
P'(x) = x(35 - x)^4 [2(35 - x) - 5x]
\]
\[
= x(35 - x)^4 (70 - 2x - 5x) = x(35 - x)^4 (70 - 7x)
\]
\[
= 7x(35 - x)^4 (10 - x)
\]

Set derivative to zero:
\[
P'(x) = 0 \Rightarrow x = 0,\ 35,\ \text{or}\ 10
\]

Check valid values:
\[ x = 0 \Rightarrow y = 35 \Rightarrow P = 0 \]
\[ x = 35 \Rightarrow y = 0 \Rightarrow P = 0 \]
\[ x = 10 \Rightarrow y = 25 \] Hence valid critical point.\[\]

Now compute second derivative to verify maximum:
\[
P''(x) = \frac{d}{dx} \left[ 7x(35 - x)^4(10 - x) \right]
\]
Use product rule:
Let \( u = x,\ v = (35 - x)^4(10 - x) \), then
\[
P''(x) = 7[u'v + uv']
\]

Compute:
\[
u' = 1,\quad v = (35 - x)^4(10 - x)
\]
\[
v' = \frac{d}{dx}[(35 - x)^4(10 - x)]
= (35 - x)^3[-4(10 - x) + (35 - x)(-1)]
\]
\[
= (35 - x)^3 [-4(10 - x) - (35 - x)] = (35 - x)^3 [-40 + 4x - 35 + x] = (35 - x)^3(-75 + 5x)
\]

So:
\[
P''(x) = 7[(35 - x)^4(10 - x) + x(35 - x)^3(-75 + 5x)]
\]

Evaluate at \( x = 10 \):
\[
(35 - 10) = 25,\quad (10 - 10) = 0\]
\[\Rightarrow P''(10) = 7[25^4 \cdot 0 + 10 \cdot 25^3(-75 + 50)] = 7 \cdot 10 \cdot 25^3(-25)
\]
\[
= -7 \cdot 10 \cdot 25^3 \cdot 25 < 0
\]

Since \( P''(10) < 0 \), the function attains a \({\bf maximum}\) at \( x = 10 \).

Thus, the required numbers are:
\[
\boxed{x = 10,\quad y = 35-x=25}
\]