Find the absolute maximum and minimum values of the function, subject to the given constraints. \[ k(x, y)=-x^{2}-y^{2}+4 x+4 y ; 0 \leq x \leq 3, y \geq 0, \text { and } x+y \leq 6 \] The minimum value of $k$ is (Simplify your answer.) The maximum value of $k$ is (Simplify your answer.)

Step 1 ：We can write the expression as \(k(x, y) = -(x^{2} - 4x + 4) - (y^{2} - 4y + 4)\)

Step 2 ：This simplifies to \(k(x, y) = -(x - 2)^{2} - (y - 2)^{2} + 4\)

Step 3 ：Since \(x\) and \(y\) are both non-negative and their sum is less than or equal to 6, the minimum value of \(x - 2\) and \(y - 2\) is -2 and the maximum is 1.

Step 4 ：Substituting these values into the equation, we find that the minimum value of \(k(x, y)\) is \(-((-2)^{2} + (-2)^{2} + 4) = -8\) and the maximum value is \(-(1^{2} + 1^{2} - 4) = 2\)

Step 5 ：Thus, the minimum value of \(k\) is \(\boxed{-8}\) and the maximum value of \(k\) is \(\boxed{2}\)