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The problem asks to find the tangent plane to the equation \( z = 6e^{x^2 - 6y} \) at the point \( (12, 24, 6) \).

The general equation of a tangent plane to a surface \( z = f(x, y) \) at the point \( (x_0, y_0, z_0) \) is given by:

\[
z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)
\]

Where:
- \( f_x(x_0, y_0) \) is the partial derivative of \( f(x, y) \) with respect to \( x \) evaluated at \( (x_0, y_0) \),
- \( f_y(x_0, y_0) \) is the partial derivative of \( f(x, y) \) with respect to \( y \) evaluated at \( (x_0, y_0) \).

### Step 1: Compute the Partial Derivatives

1. **Partial derivative with respect to \( x \)**:
\[
f_x(x, y) = \frac{\partial}{\partial x} \left( 6e^{x^2 - 6y} \right)
\]
Using the chain rule:
\[
f_x(x, y) = 6 \cdot e^{x^2 - 6y} \cdot (2x)
\]
So,
\[
f_x(x, y) = 12x \cdot e^{x^2 - 6y}
\]

2. **Partial derivative with respect to \( y \)**:
\[
f_y(x, y) = \frac{\partial}{\partial y} \left( 6e^{x^2 - 6y} \right)
\]
Again, using the chain rule:
\[
f_y(x, y) = 6 \cdot e^{x^2 - 6y} \cdot (-6)
\]
So,
\[
f_y(x, y) = -36 \cdot e^{x^2 - 6y}
\]

### Step 2: Evaluate the Partial Derivatives at \( (12, 24) \)

Now, substitute \( x = 12 \) and \( y = 24 \) into the partial derivatives.

1. **Evaluate \( f_x(12, 24) \)**:
\[
f_x(12, 24) = 12 \cdot 12 \cdot e^{12^2 - 6 \cdot 24} = 144 \cdot e^{144 - 144} = 144 \cdot e^0 = 144
\]

2. **Evaluate \( f_y(12, 24) \)**:
\[
f_y(12, 24) = -36 \cdot e^{12^2 - 6 \cdot 24} = -36 \cdot e^{144 - 144} = -36 \cdot e^0 = -36
\]

### Step 3: Write the Equation of the Tangent Plane

Using the point \( (x_0, y_0, z_0) = (12, 24, 6) \) and the values of the partial derivatives, the equation of the tangent plane is:

\[
z - 6 = 144(x - 12) - 36(y - 24)
\]

Simplify this equation:

\[
z - 6 = 144x - 1728 - 36y + 864
\]
\[
z = 144x - 36y - 1728 + 864 + 6
\]
\[
z = 144x - 36y - 858
\]

Thus, the equation of the tangent plane is:

\[
z = 144x - 36y - 858
\]

Would you like further details or have any questions? Here are some related questions:

1. How do we derive the tangent plane equation for more complex functions?
2. How do second derivatives affect the tangent plane?
3. What is the geometric interpretation of the tangent plane in 3D space?
4. How does the chain rule apply in computing partial derivatives?
5. What happens when evaluating the tangent plane at critical points?

**Tip**: The tangent plane is an excellent linear approximation of the surface near the given point!

Find the tangent plane to the equation z = 6e^(x^2 - 6y) at the point (12, 24, 6).

This page provides a detailed solution to finding the tangent plane to the equation z = 6e^(x^2 - 6y) at the point (12, 24, 6). Key multivariable calculus concepts such as partial derivatives and the tangent plane formula are applied.

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