Download Multi-variable Functions

Multi-variable Functions
MAT 1300 3X
Summer 2011
1 FUNCTIONS OF TWO VARIABLES
1
2
Functions of Two Variables
Ex: Consider a manufacturing plant which produces both car tires and shoe
soles. Let t denote the number of tires produced and let s denote the number
of shoe soles produced. Let C be the total cost in dollars of producing both
tires and soles. Hence C depends on both t and s.
C = f (t, s)
Suppose, due to physical restrictions, that at most 100 tires and 500 shoe
soles can be produced. Furthermore, suppose that the fixed cost for the plant
is $8000, the cost of producing each tire is $40, and the cost of producing
each shoe sole is $10. Find a mathematical expression for C = f (t, s) and
find the domain of f (t, s).
Ex: Compute f (20, 400) and f (100, 500).
1 FUNCTIONS OF TWO VARIABLES
3
A function of two variables, f (x, y), is a function which assigns a real value
to every ordered pair of real numbers, (x, y), which is in the domain of f .
The domain of f (x, y) is the set of all ordered pairs of real numbers, (x, y),
which can be the input for our function f (x, y).
Ex: What is the domain of
f (x, y) =
Compute the value of f (2, 1).
x ln(y + 1)
x2 + y 2
?
1 FUNCTIONS OF TWO VARIABLES
Ex: Given f (x, y) = 1 +
√
x + y + 2, find the domain of f .
4
2 FUNCTIONS OF MANY VARIABLES
2
5
Functions of Many Variables
We may have functions of more than two variables. Such functions cannot
be graphed.
Ex: Consider making a monthly contribution, M , at the end of each month,
to a GIC with an annual interest rate r, compounded monthly. The value,
V , of the GIC after T years is given by
12M r 12T
V = f (M, r, T ) =
1+
−1 .
r
12
a) Compute the value of a GIC with monthly contributions of $100 and an
annual interest rate of 4% after 30 years.
b) Compute the value of a GIC with monthly contributions of $200 and an
annual interest rate of 4% after 30 years.
3 PARTIAL DERIVATIVES
3
6
Partial Derivatives
The partial derivative of f with respect to x at the point (x0 , y0 ) is given by
∂z f (x0 + h, y0 ) − f (x0 , y0 )
fx (x0 , y0 ) =
.
= lim
∂x (x0 ,y0 ) h→0
h
The partial derivative of f with respect to y at the point (x0 , y0 ) is given by
∂z f (x0 , y0 + h) − f (x0 , y0 )
fy (x0 , y0 ) =
= lim
.
∂y (x0 ,y0 ) h→0
h
Computing Partial Derivatives:
1. To find fx (x, y) take the derivative of f with respect to x treating y as
a constant.
2. To find fy (x, y) take the derivative of f with respect to y treating x as
a constant.
3. To find either of these partial derivatives at a given point, we simply
substitute in the point after taking the partial derivative.
4. All the rules for derivation (product rule, quotient rule, chain rule) still
apply.
The geometric interpretation of partial derivatives is similar to that of the
ordinary derivative of a function of one variable. fx represents the slope of
the surface pointing in the x direction while fy represents the slope of the
surface pointing in the y direction.
3 PARTIAL DERIVATIVES
For the partial derivative of the following functions
1. f (x, y) =
p
x2 − y 2
2. f (x, y) = 4x2 y + x3 y 3 + y + 4
3. f (x, y) = e2x+3y
4. f (x, y) = exy
5. f (x, y) = x ln(xy)
7
3 PARTIAL DERIVATIVES
8
Ex: Given f (x, y) = xey + y 2 , find fx (x, y) and fy (x, y) and compute fx (2, 1)
and fy (2, 1).
3 PARTIAL DERIVATIVES
Ex: Given
x ln(x + y)
,
x2 + y 3
∂z ∂z and compute ∂x
and
.
∂x (1,1)
(1,0)
z=
find
∂z
∂x
and
∂z
∂y
9
4 HIGHER-ORDER PARTIAL DERIVATIVES
4
10
Higher-order Partial Derivatives
Since a function of two variables has two first derivatives, it has four second
derivatives. They are
fxx : take partial derivative with respect to x twice.
fyy : take partial derivative with respect to y twice.
fxy : take partial derivative with respect to x, then y.
fyx : take partial derivative with respect to y, then x.
The second-order partial derivatives of z = f (x, y) are
∂z ∂2z
∂
∂
fxx (x, y) = ∂x
= ∂x
[fx (x, y)]
= ∂x
2
∂x 2
∂ z
∂
∂z
∂
fxy (x, y) = ∂y∂x = ∂y ∂x = ∂y [fx (x, y)]
h i
∂
∂2z
∂z
∂
[fy (x, y)]
fyx (x, y) = ∂x∂y = ∂x ∂y
= ∂x
h
i
2
∂ z
∂z
∂
∂
fyy (x, y) = ∂y
= ∂y
= ∂y
[fy (x, y)] .
2
∂y
FACT - for any reasonable function, fxy = fyx .
Ex: Find all the second-order partial derivatives of
f (x, y) = xey − ln(x2 y).
4 HIGHER-ORDER PARTIAL DERIVATIVES
Ex: Find all the second-order partial derivatives of
f (x, y) = 3xy 2 + 2xy + x2 .
11
5 RELATIVE EXTREMA
5
12
Relative Extrema
A function, x = f (x, y) has a relative maximum at the point (x0 , y0 ) if for
all points (x, y) sufficiently close to (x0 , y0 ) we have that
f (x, y) ≤ f (x0 , y0 ).
A function, x = f (x, y) has a relative minimum at the point (x0 , y0 ) if for all
points (x, y) sufficiently close to (x0 , y0 ) we have that
f (x, y) ≥ f (x0 , y0 ).
For example, f (x, y) = x2 + y 2 has a relative minimum at (0, 0).
6 CRITICAL POINTS
6
13
Critical Points
Fact: Let (x0 , y0 ) be a relative extremum of f (x, y).
fy (x0 , y0 ) exist then fx (x0 , y0 ) = 0 and fy (x0 , y0 ) = 0.
If fx (x0 , y0 ) and
Any point, (x0 , y0 ) of a function f (x, y) which satisfies fx (x0 , y0 ) = 0 and
fy (x0 , y0 ) = 0 is called a critical point. Critical points are possible relative
extrema of a function.
Ex: Find all the critical points of the following functions:
1. f (x, y) = 2x2 + y 2 + 8x − 6y + 20
2. f (x, y) = −x2 − 5y 2 + 8x − 10y − 13
6 CRITICAL POINTS
3. f (x, y) = e−(x
2 +y 2 )
4. f (x, y) = x2 + 6xy + 10y 2 − 4y + 4
14
6 CRITICAL POINTS
5. f (x, y) = x3 + 2y 2 + 5xy − 1.
15
7 THE SECOND DERIVATIVE TEST
7
16
The Second Derivative Test
The Second Derivative Test for Functions of Two Variables: Suppose
that f (x, y) has continuous second-order partial derivatives at all points near
the critical point (x0 , y0 ). Let
D(x, y) = fxx (x, y)fyy (x, y) − [fxy (x, y)]2 .
1. If D(x, y) > 0 and fxx (x0 , y0 ) > 0 then (x0 , y0 ) is a relative minimum
of f (x, y).
2. If D(x, y) > 0 and fxx (x0 , y0 ) < 0 then (x0 , y0 ) is a relative maximum
of f (x, y).
3. If D(x, y) < 0 then (x0 , y0 ) is a saddle point of f (x, y).
For example, (0, 0) is a saddle point of f (x, y) = x2 − y 2 .
Picture:
7 THE SECOND DERIVATIVE TEST
Ex: Find all relative extrema of
f (x, y) = 2x2 + y 2 + 8x − 6y + 20
17
7 THE SECOND DERIVATIVE TEST
Ex: Find all relative extrema of
1
f (x, y) = x2 − 2xy + y 3 − 3y.
3
18
7 THE SECOND DERIVATIVE TEST
Ex: Find all relative extrema of
f (x, y) = x4 + 2x2 y 2 + y 2 .
19