given x^3 + y^3=6xy, At what point in the first quadrant is the tangent line horizontal?

We are given the equation:

\[
x^3 + y^3 = 6xy
\]

We are tasked with finding the point in the first quadrant where the tangent line is horizontal. A horizontal tangent line occurs when the slope of the tangent line is zero, i.e., \(\frac{dy}{dx} = 0\).

### Step 1: Implicit Differentiation
We differentiate the equation implicitly with respect to \(x\):

\[
\frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(6xy)
\]

Using the product rule for the right-hand side:

\[
3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx}
\]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearrange the equation to group terms with \(\frac{dy}{dx}\) on one side:

\[
3y^2 \frac{dy}{dx} - 6x \frac{dy}{dx} = 6y - 3x^2
\]

Factor out \(\frac{dy}{dx}\):

\[
\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2
\]

Solve for \(\frac{dy}{dx}\):

\[
\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}
\]

### Step 3: Set \(\frac{dy}{dx} = 0\)
For the tangent line to be horizontal, \(\frac{dy}{dx} = 0\). This occurs when the numerator of the derivative is zero:

\[
6y - 3x^2 = 0
\]

Simplify:

\[
2y = x^2
\]

### Step 4: Substitute \(y = \frac{x^2}{2}\) into the original equation
Now, substitute \(y = \frac{x^2}{2}\) into the original equation \(x^3 + y^3 = 6xy\):

\[
x^3 + \left(\frac{x^2}{2}\right)^3 = 6x\left(\frac{x^2}{2}\right)
\]

Simplify each term:

\[
x^3 + \frac{x^6}{8} = 3x^3
\]

Multiply through by 8 to eliminate the fraction:

\[
8x^3 + x^6 = 24x^3
\]

Rearrange the equation:

\[
x^6 - 16x^3 = 0
\]

Factor:

\[
x^3(x^3 - 16) = 0
\]

This gives us two solutions:

\[
x^3 = 0 \quad \text{or} \quad x^3 = 16
\]

Since we are looking for a point in the first quadrant, \(x^3 = 0\) does not provide a valid solution. Therefore:

\[
x^3 = 16 \quad \Rightarrow \quad x = \sqrt[3]{16}
\]

### Step 5: Solve for \(y\)
Substitute \(x = \sqrt[3]{16}\) into \(y = \frac{x^2}{2}\):

\[
y = \frac{\left(\sqrt[3]{16}\right)^2}{2} = \frac{\sqrt[3]{256}}{2}
\]

### Final Answer:
The point in the first quadrant where the tangent line is horizontal is:

\[
\left( \sqrt[3]{16}, \frac{\sqrt[3]{256}}{2} \right)
\]

Would you like further details or explanations on any part of this solution? 

Here are five questions related to this problem:
1. How is implicit differentiation applied to other types of equations?
2. Why does a horizontal tangent imply \(\frac{dy}{dx} = 0\)?
3. What does the product rule in differentiation represent geometrically?
4. How can this approach be extended to find vertical tangents?
5. What does the solution tell us about the geometry of the curve?

**Tip:** When solving equations involving products, implicit differentiation and the product rule are crucial for finding derivatives efficiently.

given x^3 + y^3 = 6xy, At what point in the first quadrant is the tangent line horizontal?

Discover how to find the point in the first quadrant where the tangent line is horizontal for the equation x^3 + y^3 = 6xy. This step-by-step solution uses implicit differentiation, product rule, and algebraic factorization. Ideal for Grades 11-12 students.

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