330 CHAPTER 18. OPTIMIZATION

Theorem 18.1.3 Suppose U is an open set contained in D( f ) such that f is differentiableon U and suppose x ∈U is a local minimum or local maximum for f . Then ∇ f (x) = 0.

Definition 18.1.4 A singular point for f is a point x where ∇ f (x) = 0. This is also calleda critical point. By analogy with the one variable case, a point where the gradient does notexist will also be called a critical point.

Example 18.1.5 Find the critical points for the function f (x,y)≡ xy− x− y for x,y > 0.

Note that here D( f ) is an open set and so every point is an interior point. Where is thegradient equal to zero? fx = y−1 = 0, fy = x−1 = 0, and so there is exactly one criticalpoint (1,1).

Example 18.1.6 Find the volume of the smallest tetrahedron made up of the coordinateplanes in the first octant and a plane which is tangent to the sphere x2 + y2 + z2 = 4.

The normal to the sphere at a point (x0,y0,z0) is(

x0,y0,√

4− x20− y2

0

)and so the

equation of the tangent plane at this point is

x0 (x− x0)+ y0 (y− y0)+√

4− x20− y2

0

(z−√

4− x20− y2

0

)= 0

When x = y = 0, z = 4√(4−x2

0−y20)

. When z = 0 = y, x = 4x0

, and when z = x = 0, y = 4y0

.

Therefore, the function to minimize is

f (x,y) =16

64

xy√(4− x2− y2)

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

−4+2x2 + y2

x2y(−4+ x2 + y2)√(4− x2− y2)

= 0,

and−4+ x2 +2y2

xy2 (−4+ x2 + y2)√(4− x2− y2)

= 0.

Therefore, x2 +2y2 = 4 and 2x2 + y2 = 4. Thus x = y and so x = y = 2√3. It follows from

the equation for z that z = 2√3

also. How do you know this is not the largest tetrahedron?

Example 18.1.7 An open box is to contain 32 cubic feet. Find the dimensions which willresult in the least surface area.

Let the height of the box be z and the length and width be x and y respectively. Thenxyz = 32 and so z = 32/xy. The total area is xy+ 2xz+ 2yz and so in terms of the twovariables x and y, the area is A = xy+ 64

y + 64x . To find best dimensions you note these must

result in a local minimum.

Ax =yx2−64

x2 = 0, Ay =xy2−64

y2 .