To find the maximum value of the function $f(x, y) = x^3 y^9$ for $x, y \geq 0$ on the unit circle, we start with the constraint given by the unit circle:

$x^2 + y^2 = 1.$

Step 1: Use the constraint

We can express $y$ in terms of $x$ using the constraint:

$y = \sqrt{1 - x^2}.$

Step 2: Substitute into the function

Now, substitute $y$ into the function:

$f(x) = x^3 (\sqrt{1 - x^2})^9 = x^3 (1 - x^2)^{\frac{9}{2}}.$

Step 3: Find the derivative

To find the critical points, we differentiate $f(x)$:

1. Using the product rule:

Let $u = x^3$ and $v = (1 - x^2)^{\frac{9}{2}}$.

The derivative $f'(x)$ is given by:

$f'(x) = u'v + uv'.$

Calculating $u'$ and $v'$:

$u' = 3x^2,$ $v' = \frac{9}{2} (1 - x^2)^{\frac{7}{2}} (-2x) = -9x(1 - x^2)^{\frac{7}{2}}.$

Thus,

$f'(x) = 3x^2 (1 - x^2)^{\frac{9}{2}} - 9x^4 (1 - x^2)^{\frac{7}{2}}.$

Step 4: Set the derivative to zero

Setting $f'(x) = 0$:

$3x^2 (1 - x^2)^{\frac{7}{2}} (1 - 3x^2) = 0.$

This gives us two cases:

1. $x^2 = 0$ (implying $x = 0$).
2. $1 - 3x^2 = 0$ (implying $x^2 = \frac{1}{3}$ or $x = \frac{1}{\sqrt{3}}$).

Step 5: Calculate $y$

For $x = \frac{1}{\sqrt{3}}$:

$y = \sqrt{1 - \left( \frac{1}{\sqrt{3}} \right)^2} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}}.$

Step 6: Evaluate $f(x, y)$

Now evaluate $f$ at the critical points:

1. At $(0, 1)$:

$f(0, 1) = 0^3 \cdot 1^9 = 0.$

2. At $\left( \frac{1}{\sqrt{3}}, \sqrt{\frac{2}{3}} \right)$:

$f\left( \frac{1}{\sqrt{3}}, \sqrt{\frac{2}{3}} \right) = \left( \frac{1}{\sqrt{3}} \right)^3 \left( \sqrt{\frac{2}{3}} \right)^9.$

Calculating:

$= \frac{1}{3\sqrt{3}} \cdot \left( \frac{2^4}{3^4} \cdot \sqrt{2} \right) = \frac{16\sqrt{2}}{81\sqrt{3}}.$

Step 7: Check boundary points

Check $(1, 0)$ and $(0, 1)$:

• $f(1, 0) = 1^3 \cdot 0^9 = 0$.
• $f(0, 1) = 0$.

Conclusion

The maximum occurs at $\left( \frac{1}{\sqrt{3}}, \sqrt{\frac{2}{3}} \right)$:

$f\left( \frac{1}{\sqrt{3}}, \sqrt{\frac{2}{3}} \right) = \frac{16\sqrt{2}}{81\sqrt{3}}.$

Thus, the maximum value of $f(x, y)$ on the unit circle is:

$\boxed{\frac{16\sqrt{2}}{81\sqrt{3}}}.$

If you have any questions or need further details, let me know! Here are some related questions you might consider:

1. How does the method of Lagrange multipliers apply to this problem?
2. What are the geometric interpretations of the critical points found?
3. How does this function behave as $x$ and $y$ approach the unit circle's boundary?
4. Can we find the maximum value using a different method?
5. How would the maximum change if we altered the constraint?

Tip: Always check boundary points in optimization problems with constraints!