Let #f(x,y)=8x^3+y^3-6x-12y+4#, how do you determine all critical points?

Question

Cameron Alexander · Accepted Answer

Find the first partial derivatives.
Set them equal to #0#.
Solve the system of equations. #f_x = 24x^2-6=0#
#f_y = 3y^2-12=0#

Emma Aldridge · Answer

undefined To determine all critical points of the function $ f(x, y) = 8x^3 + y^3 - 6x - 12y + 4 $, we need to find where the partial derivatives with respect to $ x $ and $ y $ are both zero.

1. Compute the partial derivatives:
   $$
   \frac{\partial f}{\partial x} = 24x^2 - 6 \quad \text{and} \quad \frac{\partial f}{\partial y} = 3y^2 - 12
   $$

2. Set each partial derivative equal to zero and solve for $ x $ and $ y $:
   $$
   \frac{\partial f}{\partial x} = 0 \Rightarrow 24x^2 - 6 = 0 \Rightarrow x^2 = \frac{1}{4} \Rightarrow x = \pm \frac{1}{2}
   $$
   $$
   \frac{\partial f}{\partial y} = 0 \Rightarrow 3y^2 - 12 = 0 \Rightarrow y^2 = 4 \Rightarrow y = \pm 2
   $$

3. Combine the values of $ x $ and $ y $ to get the critical points:
   $$
   (x, y) = \left(\frac{1}{2}, 2\right), \left(\frac{1}{2}, -2\right), \left(-\frac{1}{2}, 2\right), \left(-\frac{1}{2}, -2\right)
   $$

Therefore, the critical points of the function are $ \left(\frac{1}{2}, 2\right) $, $ \left(\frac{1}{2}, -2\right) $, $ \left(-\frac{1}{2}, 2\right) $, and $ \left(-\frac{1}{2}, -2\right) $.