The following definition describes what is meant by a local maximum or local minimum.
Definition 13.1.1Suppose f : D
(f)
→ ℝ where D
(f)
⊆ ℝ^{n}. A point x ∈ D
(f)
⊆ ℝ^{n}is called alocal minimum if f
(x)
≤ f
(y )
for all y ∈ D
(f)
sufficiently close tox.A point x ∈ D
(f)
is calledalocal maximum if f
(x)
≥ f
(y)
for all y ∈ D
(f)
sufficiently close tox.Alocal extremum isa point of D
(f)
which is either a local minimum or a local maximum. The plural for extremum isextrema. The plural for minimum is minima and the plural for maximum is maxima.
Procedure 13.1.2To find candidates for local extrema which are interior points of D
(f)
where fis a differentiable function, you simply identify those points where ∇f equals the zero vector.
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
(t)
≡ f
(x + tv )
for small
|t|
. Since f has a local maximum (minimum), it
follows that h is a differentiable function of the single variable t for small t which has a local maximum
(minimum) when t = 0. Therefore, h^{′}
(0)
= 0.
h (Δt )− h(0) = f (x + Δtv)− f (x)
= Df (x)Δtv + o(Δt)
Now divide by Δt and let Δt → 0 to obtain
0 = h′(0) = Df (x )v
and since v is arbitrary, it follows Df
(x)
= 0. However,
( )
Df (x) = fx1 (x) ⋅⋅⋅ fxn (x)
and so ∇f
(x)
= 0. This proves the following theorem.
Theorem 13.1.3Suppose U is an open set contained in D
(f)
such that f is differentiable on Uand suppose x ∈ U is a local minimum or local maximum for f. Then ∇f
(x)
= 0.
Definition 13.1.4A singular point for f is a point x where ∇f
(x)
= 0.This is also called acritical point. By analogy with the one variable case, a point where the gradient does not exist willalso be called a critical point.
Example 13.1.5Find 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 the
gradient equal to zero? f_{x} = y − 1 = 0,f_{y} = x − 1 = 0, and so there is exactly one critical point
(1,1)
.
Example 13.1.6Find the volume of the smallest tetrahedron made up of the coordinate planes inthe 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
(x0,y0,z0)
of the sphere is
( ∘-----2---2)
x0,y0, 4 − x0 − y0
and so the
equation of the tangent plane at this point is
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
Therefore, x^{2} + 2y^{2} = 4 and 2x^{2} + y^{2} = 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 13.1.7An open box is to contain 32 cubic feet. Find the dimensions which will result inthe 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 +
64-
y
+
64
x
. To find best dimensions you note these must result in a local
minimum.
A = yx2-−-64-= 0,A = xy2 −-64-.
x x2 y y2
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?