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To find the critical points of the function \( f(x, y) = 7 + x - 6x^2 - 4y + 2y^2 \), we need to compute the partial derivatives with respect to \(x\) and \(y\), set them equal to zero, and solve the resulting system of equations.

### Step 1: Compute the partial derivatives.
- The partial derivative with respect to \(x\) is:
  \[
  f_x = \frac{\partial}{\partial x}(7 + x - 6x^2 - 4y + 2y^2) = 1 - 12x.
  \]

- The partial derivative with respect to \(y\) is:
  \[
  f_y = \frac{\partial}{\partial y}(7 + x - 6x^2 - 4y + 2y^2) = -4 + 4y.
  \]

### Step 2: Set the partial derivatives equal to zero.
- \( f_x = 1 - 12x = 0 \) implies:
  \[
  12x = 1 \implies x = \frac{1}{12}.
  \]

- \( f_y = -4 + 4y = 0 \) implies:
  \[
  4y = 4 \implies y = 1.
  \]

### Step 3: The critical point.
The critical point of the function is \( \left( \frac{1}{12}, 1 \right) \).

Would you like more details on verifying the nature of this critical point (e.g., using the second derivative test)?

### Related Questions:
1. How do you verify whether a critical point is a maximum, minimum, or saddle point?
2. What is the significance of finding partial derivatives in a function of multiple variables?
3. How would the critical points change if the coefficients in the function were different?
4. How do you interpret the meaning of a critical point in a real-world context?
5. What role do second derivatives play in determining the nature of a critical point?

### Tip:
When finding critical points for functions of two variables, setting the partial derivatives to zero helps locate points where the function's slope is flat in both directions. This technique is fundamental in optimization problems.

Find the critical point of the function: f(x, y) = 7 + x - 6x^2 - 4y + 2y^2.

Learn how to find the critical points of the function f(x, y) = 7 + x - 6x^2 - 4y + 2y^2 using partial derivatives. This problem covers key concepts in multivariable calculus, including the first derivative test.

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