Find all critical points of the following function. \( f(x, y)=x y+x^{4}+y^{4} \) What are the critical points? Select the correct choice below and, if necessary, fill in the answer box within your choice. A. The critical point(s) is/are (Type an ordered pair. Use a comma to separate answers as needed.) B. There are no critical points.

To find the critical points of the function \( f(x, y) = xy + x^4 + y^4 \), we first need to compute the partial derivatives and set them equal to zero. 1. Compute the partial derivatives: - \( f_x = \frac{\partial f}{\partial x} = y + 4x^3 \) - \( f_y = \frac{\partial f}{\partial y} = x + 4y^3 \) 2. Set the partial derivatives equal to zero: - \( y + 4x^3 = 0 \) (1) - \( x + 4y^3 = 0 \) (2) 3. From equation (1), we can solve for \( y \): \[ y = -4x^3 \] 4. Substitute this expression for \( y \) into equation (2): \[ x + 4(-4x^3)^3 = 0 \] \[ x + 4(-64x^9) = 0 \] \[ x - 256x^9 = 0 \] \[ x(1 - 256x^8) = 0 \] This gives us two cases: - Case 1: \( x = 0 \) - Case 2: \( 1 - 256x^8 = 0 \Rightarrow 256x^8 = 1 \Rightarrow x = \pm \frac{1}{2} \) Next, we will calculate the corresponding \( y \) values for each \( x \): - If \( x = 0 \), substituting in equation (1) gives \( y = 0 \). So, one critical point is \( (0, 0) \). - If \( x = \frac{1}{2} \): \[ y = -4\left(\frac{1}{2}\right)^3 = -4 \times \frac{1}{8} = -\frac{1}{2} \] Thus, another critical point is \( \left(\frac{1}{2}, -\frac{1}{2}\right) \). - If \( x = -\frac{1}{2} \): \[ y = -4\left(-\frac{1}{2}\right)^3 = -4 \times -\frac{1}{8} = \frac{1}{2} \] Hence, a third critical point is \( \left(-\frac{1}{2}, \frac{1}{2}\right) \). The critical points are: A. The critical point(s) is/are (0, 0), (1/2, -1/2), (-1/2, 1/2).