Find the absolute maximum and minimum of $f(x, y)=5 x+4 y$ within the domain $x^{2}+y^{2} \leq 25$. 1. Absolute minimum of $f(x, y)$ is 2. Absolute maximum of $f(x, y)$ is

Step 1 ：We are given the function \(f(x, y)=5x+4y\) and we are asked to find the absolute maximum and minimum within the domain \(x^{2}+y^{2} \leq 25\).

Step 2 ：Since \(f(x, y)\) is a linear function, it has no critical points. Therefore, the absolute maximum and minimum must occur on the boundary of the domain, which is the circle \(x^2 + y^2 = 25\).

Step 3 ：We can use the method of Lagrange multipliers to find the maximum and minimum of \(f(x, y)\) on the boundary. The Lagrangian is \(L(x, y, \lambda) = 5x + 4y - \lambda(x^2 + y^2 - 25)\), and we need to solve the system of equations \(\nabla L = 0\).

Step 4 ：Solving the system of equations, we get two solutions \((-25\sqrt{41}/41, -20\sqrt{41}/41, -\sqrt{41}/10)\) and \((25\sqrt{41}/41, 20\sqrt{41}/41, \sqrt{41}/10)\).

Step 5 ：Substituting these solutions into the function \(f(x, y)\), we get the values \(-5\sqrt{41}\) and \(5\sqrt{41}\).

Step 6 ：Therefore, the absolute minimum of \(f(x, y)\) is \(\boxed{-5\sqrt{41}}\) and the absolute maximum of \(f(x, y)\) is \(\boxed{5\sqrt{41}}\).