 So, if we want to find an extreme value of some objective function subject to some constraint, we can solve the system of equations where the gradient of the objective is a scalar multiple of the gradient of the constraint, and the natural question to ask is, will this always work? No. And here's the problem. The basic requirement that the objective function and constraint function are cotangent, well, they can only meet at one point. That still holds. However, they don't have to be cotangent at that point. In fact, one or both tangent lines might not exist. And this can happen in two ways. If our functions are discontinuous, then that's really a horrifying nightmare. So if our functions are continuous, and we really want them to be continuous or the problem becomes much harder, the tangent lines will fail to exist at cusps. So remember the gradient corresponds to the vector normal to the level curve, f of x, y equals c. And this will be normal to the tangent line. So if we're at a cusp, there's two possibilities. First, the tangent line might not exist at all. And in that case, our gradient would be undefined. The other possibility at a cusp is that the tangent line might be vertical. And in that case, the gradient is going to be equal to the zero vector. And these conditions should look very familiar, because remember when you were finding optimal values for functions of a single variable, the critical points were places where the derivative was zero or failed to exist. And what this means is that in addition to the solutions by the Lagrangian, you should also check points where the gradient is the zero vector or undefined. For example, let's try to find the extreme values of x on the curve y squared equals x cubed. So the objective function, well, we want to find the extreme value of x. So our objective function is L of x, y equals x. So our partial derivatives are, we are on the curve y squared equals x cubed. And so we'll make our constraint function x cubed minus y squared, the partial derivatives are. We want the gradient of L to be lambda times the gradient of f, comparing the components. Now, from the second equation, we have either y equals zero or lambda equals zero. But if lambda equals zero, 3x squared lambda can't equal one. And if y equals zero, since we're on the curve y squared equals x cubed, then x equals zero. But if x equals zero, 3x squared lambda can't equal one. And so the system has no solution. However, we note that the gradient of f is the vector 3x squared minus 2y. And so the gradient will be the zero vector at x equals zero, y equals zero. Now, here's an important thing to check. This actually is a point on y squared equals x cubed. So it is, in fact, a feasible value. And it corresponds to the minimum value of x. You should take a moment to convince yourself that this is, in fact, the case.