QPaperGen

We are given:
\[
x + y = 60 \Rightarrow y = 60 - x
\]

We want to maximize:
\[
f(x) = x y^3 = x(60 - x)^3
\]

Differentiate using the product rule:
\[
f'(x) = \frac{d}{dx} \left[x(60 - x)^3\right] = (60 - x)^3 + x \cdot 3(60 - x)^2 \cdot (-1)
\]
\[
f'(x) = (60 - x)^2 \left[(60 - x) - 3x\right] = (60 - x)^2(60 - 4x)
\]

Set derivative to zero:
\[
f'(x) = 0 \Rightarrow (60 - x)^2(60 - 4x) = 0
\]
\[
\Rightarrow x = 60 \text{ or } x = 15
\]

Check second derivative at \( x = 15 \):

\[
f''(x) = \frac{d}{dx}\left[ (60 - x)^2(60 - 4x) \right]
\]
Using the product rule:
\[
f''(x) = 2(60 - x)(-1)(60 - 4x) + (60 - x)^2(-4)
\]
\[
f''(x) = -2(60 - x)(60 - 4x) - 4(60 - x)^2
\]
\[
f''(x) = -2(60 - x)[(60 - 4x) + 2(60 - x)] = -2(60 - x)(60 - 4x + 2(60 - x))
\]
\[
= -2(60 - x)(180 - 6x) = -12(60 - x)(30 - x)
\]

Now evaluate at \( x = 15 \):
\[
f''(15) = -12(60 - 15)(30 - 15) = -12(45)(15) < 0
\]

Since \( f''(15) < 0 \), \( x = 15 \) is a point of **local maximum**.

Hence,
\[
x = 15, \quad y = 60 - 15 = 45
\]



The two positive numbers are 15 and 45.