Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1

I assume there are some plus signs that aren't rendering for some reason, so that the plane should be .

You're minimizing subject to the constraint . Note that and attain their extrema at the same values of , so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is



Take your partial derivatives and set them equal to 0:



Adding the first three equations together yields



and plugging this into the first three equations, you find a critical point at .

The squared distance is then , which means the shortest distance must be .