Section 14.8: Lagrange Multipliers

SOLUTION According to the procedure of Lagrange Multipliers, we compare the values of \(f\) at the critical points with values at the points on the boundary. Since \(f_x = 2x\) and \(f_y = 4y\), the only critical point is \((0, 0)\). We compare the value of \(f\) at that point with the extreme values on the boundary from Example 2: \[ f(0, 0) = 0 \quad f(\pm 1, 0) = 1 \quad f(0, \pm 1) = 2 \] Therefore the maximum value of \(f\) on the disk \(x^2 + y^2 \le 1\) is \(f(0, \pm 1) = 2\) and the minimum value is \(f(0, 0) = 0\).