Find the maximum value of f(x,y)=x3y9 for x,y≥0 on the unit circle.

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!

Find the maximum value of f(x,y)=x^3y^9 for x,y≥0 on the unit circle.

Explore how to find the maximum value of f(x,y) = x^3y^9 subject to the constraint of the unit circle (x^2 + y^2 = 1). The detailed solution includes the use of calculus, critical points, and evaluation at boundary points, resulting in a maximum value of 16√2/(81√3). Suitable for advanced high school students.

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