Section 14.7: Max Min Problems

SOLUTION The distance from any point \((x, y, z)\) to the point \((1, 0, -2)\) is \[ d = \sqrt{(x - 1)^2 + y^2 + (z + 2)^2} \] but if \((x, y, z)\) lies on the plane \(x + 2y + z = 4\), then \(z = 4 - x - 2y\) and so we have \(d = \sqrt{(x - 1)^2 + y^2 + (6 - x - 2y)^2}\). We can minimize \(d\) by minimizing the simpler expression \[ d^2 = f(x, y) = (x - 1)^2 + y^2 + (6 - x - 2y)^2 \] By solving the equations \[ f_x = 2(x - 1) - 2(6 - x - 2y) = 4x + 4y - 14 = 0 \] \[ f_y = 2y - 4(6 - x - 2y) = 4x + 10y - 24 = 0 \] we find that the only critical point is \((11/6, 5/3)\). Since \(f_{xx} = 4\), \(f_{xy} = 4\), and \(f_{yy} = 10\), we have \(D(x, y) = f_{xx}f_{yy} - (f_{xy})^2 = 24 > 0\) and \(f_{xx} > 0\), so by the Second Derivatives Test \(f\) has a local minimum at \((11/6, 5/3)\). Intuitively, we can see that this local minimum is actually an absolute minimum because there must be a point on the given plane that is closest to \((1, 0, -2)\). If \(x = 11/6\) and \(y = 5/3\), then \[ d = \sqrt{(11/6 - 1)^2 + (5/3)^2 + (6 - 11/6 - 10/3)^2} = \sqrt{(5/6)^2 + (10/6)^2 + (5/6)^2} = \frac{\sqrt{150}}{6} = \frac{5\sqrt{6}}{6} \] The shortest distance from \((1, 0, -2)\) to the plane \(x + 2y + z = 4\) is \(5\sqrt{6}/6\).