Question

Find any critical points and relative extrema of the function. $$ f(x, y)=x^{2}+y^{2}+2 x-6 y+6 $$

Short answer

The function has one critical point at (-1,3), and this point is a relative minimum.

Step-by-step solution

Find the Gradient of the Function

The gradient of the function, which is the vector of all first partial derivatives, is found by taking the derivative of \(f(x, y)\) with respect to each variable. For \(f(x, y)=x^{2}+y^{2}+2 x-6 y+6\), the gradient of f is \(\nabla f = (2x + 2, 2y - 6)\)

Find the Critical Points

Now set the gradient obtained in the previous step to zero and solve for x and y. This gives us the equations \(2x + 2 = 0\) and \(2y - 6 = 0\). Solving these gives \(x = -1\) and \(y = 3\), so there's one critical point at (-1, 3).

Use the Second Derivative Test

The second derivative test determines whether the critical point is a maximum, a minimum, or a saddle point. It involves finding the Hessian matrix (the matrix of all second partial derivatives) and calculating its determinant. The second-order partial derivatives of \(f(x, y)\) are all constant: \(f_{xx} = 2\), \(f_{xy} = f_{yx} = 0\), \(f_{yy} = 2\). The determinant of the Hessian matrix, often called 'D', is \(D = f_{xx} f_{yy} - (f_{xy})^2 = (2)(2) - (0)^2 = 4\). Because \(D > 0\) and \(f_{xx} > 0\), the point (-1,3) is a relative minimum.