Question

Find all values of \(x\) and \(y\) such that \(f_{x}(x, y)=0\) and \(f_{y}(x, y)=0\) simultaneously. $$ f(x, y)=3 x^{3}-12 x y+y^{3} $$

Short answer

The solutions are \(x = \pm 2\) and \(y = \mp 2\sqrt{3}\)

Step-by-step solution

Compute the partial derivatives

In this step, derive \(f(x, y)\) partially with respect to \(x\) and \(y\). Let's denote the partial derivative with respect to \(x\) as \(f_{x}(x, y)\) and the one with respect to \(y\) as \(f_{y}(x, y)\). For \(f_{x}(x, y)\), differentiate \(f(x, y)\) with respect to \(x\), treating \(y\) as a constant. Likewise, to find \(f_{y}(x, y)\), differentiate \(f(x, y)\) with respect to \(y\), treating \(x\) as a constant. The partial derivatives are therefore: \[ f_{x}(x, y) = 9x^{2} - 12y, \] and \[ f_{y}(x, y) = -12x + 3y^{2}.\]

Set the partial derivatives equal to zero

For the critical points, both \(f_{x}(x, y)\) and \(f_{y}(x, y)\) have to be equal to zero. This gives us the following equations: \[ 9x^{2} - 12y = 0, \] and \[ -12x + 3y^{2} = 0.\] These simultaneous equations will be solved for \(x\) and \(y\).

Solve for \(x\) and \(y\)

To solve for \(x\) and \(y\), we can manipulate the first equation to give \(x\) in terms of \(y\): \(x^{2} = \frac{4}{3}y\). Substituting into the second equation, we get \(y = \pm 2\sqrt{3}x\). Hence the solutions of the system of equations are \( (\pm 2, \mp 2\sqrt{3}) .\)