Question

Find values of \(x\) and \(y\) such that \(f_{x}(x, y)=0\) and \(f_{y}(x, y)=0\) simultaneously. $$ f(x, y)=x^{2}+4 x y+y^{2}-4 x+16 y+3 $$

Short answer

The values of \(x\) and \(y\) that satisfy \(f_{x}(x, y)=0\) and \(f_{y}(x, y)=0\) simultaneously are \(x = -2\) and \(y = -3\).

Step-by-step solution

Differentiate \(f(x, y)\) with respect to \(x\)

The partial derivative of \(f\) with respect to \(x\) is: \(f_{x}(x, y) = 2x + 4y - 4\). Set this equal to 0 to get the equation \(2x + 4y - 4 = 0\).

Differentiate \(f(x, y)\) with respect to \(y\)

The partial derivative of \(f\) with respect to \(y\) is: \(f_{y}(x, y) = 4x + 2y + 16\). Set this equal to 0 to get the equation \(4x + 2y + 16 = 0\).

Solve the system of equations

The last step is to solve the system of equations \(2x + 4y - 4 = 0\) and \(4x + 2y + 16 = 0\) simultaneously. This yields solutions for \(x\) and \(y\), which are \(x = -2\) and \(y = -3\).