The following definition describes what is meant by a local maximum or local minimum.
Definition 22.1.1 Suppose f : D
Procedure 22.1.2 To find candidates for local extrema which are interior points of D
To locate candidates for local extrema, for the function f, take ∇f and find where this vector equals 0.
Let v be any vector in ℝ^{n} and suppose x is a local maximum (minimum) for f. Then consider the real valued function of one variable, h

and since v is arbitrary, it follows Df

and so ∇f
Theorem 22.1.3 Suppose U is an open set contained in D
Definition 22.1.4 A singular point for f is a point x where ∇f
Example 22.1.5 Find the critical points for the function f
Note that here D
Example 22.1.6 Find the volume of the smallest tetrahedron made up of the coordinate planes in the first octant and a plane which is tangent to the sphere x^{2} + y^{2} + z^{2} = 4.
The normal to the sphere at a point

and so the equation of the tangent plane at this point is

When x = y = 0,z =

This is because in beginning calculus it was shown that the volume of a pyramid is 1/3 the area of the base times the height. Therefore, you simply need to find the gradient of this and set it equal to zero. Thus upon taking the partial derivatives, you need to have

and

Therefore, x^{2} + 2y^{2} = 4 and 2x^{2} + y^{2} = 4. Thus x = y and so x = y =
Example 22.1.7 An open box is to contain 32 cubic feet. Find the dimensions which will result in the least surface area.
Let the height of the box be z and the length and width be x and y respectively. Then xyz = 32 and so z = 32∕xy. The total area is xy + 2xz + 2yz and so in terms of the two variables x and y, the area is A = xy +

Therefore, yx^{2} − 64 = 0 and xy^{2} − 64 = 0 so xy^{2} = yx^{2}. For sure the answer excludes the case where any of the variables equals zero. Therefore, x = y and so x = 4 = y. Then z = 2 from the requirement that xyz = 32. How do you know this gives the least surface area? Why is this not the largest surface area?