Find the extremum of $f(x, y)$ subject to the given constraint, and state whether it is a maximum or a minimum.
\[
f(x, y)=40-x^{2}-y^{2} ; x+6 y=37
\]
There is a value of located at $(x, y)=$ (Simplify your answers.)

Step 1 ：Define the function and the constraint: \(f(x, y) = 40 - x^{2} - y^{2}\) and \(g(x, y) = x + 6y - 37\).

Step 2 ：Set up the Lagrangian function: \(L = -\lambda(x + 6y - 37) - x^{2} - y^{2} + 40\).

Step 3 ：Take the partial derivatives of the Lagrangian with respect to each variable and set them equal to zero: \(L_{x} = -\lambda - 2x = 0\), \(L_{y} = -6\lambda - 2y = 0\), and \(L_{\lambda} = -x - 6y + 37 = 0\).

Step 4 ：Solve the system of equations to find the values of the variables: \(x = 1\), \(y = 6\), and \(\lambda = -2\).

Step 5 ：Check the second order conditions by calculating the determinant of the Hessian matrix of the Lagrangian and the second derivative of the Lagrangian with respect to x and y: \(L_{xx} = -2\), \(L_{xy} = 0\), \(L_{yx} = 0\), \(L_{yy} = -2\), and the determinant of the Hessian matrix is 4.

Step 6 ：Since the determinant of the Hessian matrix is positive and the second derivative of the Lagrangian with respect to x and y is negative, the point \((x, y) = (1, 6)\) is a maximum of the function \(f(x, y)\) subject to the constraint \(g(x, y) = 0\).

Step 7 ：Final Answer: The extremum of the function \(f(x, y) = 40 - x^{2} - y^{2}\) subject to the constraint \(x + 6y = 37\) is a maximum located at \((x, y) = \boxed{(1, 6)}\).