Use Lagrange multipliers to find the maximum and minimum values of $f(x, y)=x y$
subject to the constraint
\[
4 x+3 y=144
\]
if such values exist.
Enter the exact answer. If there is no global maximum or global minimum, enter NA.
Optimal $f(x, y)=$

Step 1 ：We can consider \(f(x, y) = xy\) as the product of \(x\) and \(y\). Unfortunately, their sum is not constant.

Step 2 ：In order to obtain a constant sum, we consider \((4x)(3y)\). By AM-GM,

Step 3 ：\[\sqrt[2]{(4x)(3y)} \le \frac{4x + 3y}{2} = \frac{144}{2} = 72,\]

Step 4 ：so \((4x)(3y) \le 5184\). Then

Step 5 ：\[f(x, y) = xy \le 432.\]

Step 6 ：Equality occurs when \(4x = 3y\). We can solve to get \(x = 36\) and \(y = 48\), so the maximum value is \(\boxed{432}\).

Step 7 ：Since the function \(f(x, y) = xy\) is continuous and the constraint \(4x + 3y = 144\) is a closed and bounded set, the function \(f(x, y)\) must attain a minimum value on this set. However, the minimum value is not attained at the critical point \((36, 48)\), so it must be attained at the boundary points where \(x = 0\) or \(y = 0\).

Step 8 ：When \(x = 0\), \(f(x, y) = 0\). When \(y = 0\), \(f(x, y) = 0\). So the minimum value is \(\boxed{0}\).