Partial CalculusofSeveralVariablesand
ItsApplications
Revision Derivatives
Applied Maxima & Minima.
Lagrange Multipliers.
FirstPartialDerivatives
Suppose that z = f(x, y),
The first partial derivative of f with respect to x at a
z
point (x, y) with y treated as a constant is x .
The first partial derivative of f with repsect to y at a
z
point (x, y) with x treated as a constant is
.
y
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Problem1
Find the first partial derivatives for each functions:
i) f ( x, y ) = 4 x 2 + 5 x 2 y 2 + 6 y 3 7
ii) z = x 2 y + e x ln( xy + 1)
(
iii) R = st + t
3
)
2
Solution:
i) f x ( x, y ) = 4( 2 x ) + 5( 2 x ) y 2 = 8 x + 10 xy 2
f y ( x, y ) = 5 x 2 ( 2 y ) + 6( 3 x 2 ) = 10 x 2 y + 18 y 2
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Continued...
ii) z = ( 2 x ) y + e x y = 2 xy + e x y
x
xy + 1
xy + 1
z
x
x
2
2
= x (1) + 0
=x
y
xy + 1
xy + 1
)[
]
iii) R
= 2 st 3 + t (1) t 3 = 2t 3 st 3 + t
s
(
(
)[( ) ] (
)
R
= 2 st 3 + t s 3t 2 + 1 = 2 st 3 + t 3st 2 + 1
t
(
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)
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SecondOrderPartialDerivatives
The second partial derivatives of a function f
with two variables, x and y, are fxx, fyy, fxy, and fyx
2
where f
f xx = 2 = ( f x )
x
x
2 f
f yy = 2 = ( f y )
y
y
2 f
f xy =
= ( fx )
yx y
2 f
f yx =
= ( fy )
xy x
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Problem2
zof x 2 y + e xy .
Find each of the second partial derivatives =
Solution:
First partial derivatives:
z
= ( 2 x ) y + e xy ( y ) = 2 xy + ye xy
x
z
= x 2 (1) + e xy ( x ) = x 2 + xe xy
y
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Continued...
Thus, the second partial
derivatives:
2z
= 2(1) y + ye xy ( y ) = 2 y + y 2 e xy
x 2
2z
= 0 + xe xy x = x 2 e xy
y 2
[(
]
2z
= 2 x(1) + y e xy x + e xy (1) = 2 x + xye xy + e xy
yx
2z
= 2 x + x e xy y + e xy (1) = 2 x + xye xy + e xy
xy
)
[(
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ApplicationsofFirst
PartialDerivatives
An important problem in economics concerns how
the factors necessary for a production determine
the output of a product.
For example, the output of a product depends on
labor, land, capital, material and machines.
If the amount of output z of a product depends on
the amounts of two inputs x and y, then the
quantity z is given by the production (function
z = f x, y ).
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ApplicationsofFirst
PartialDerivatives
z
x
represents the rate of change in the output z
with resepect to input x while input y remains
constant. This partial derivative is called the
z
marginal productivity y x. The partial derivative
of
is the marginal prodcutivity of y and measures the
rate of change of z with respect to input y.
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Example
Suppose that the production function for a product is
, where x represents the number
z = x ln ( y + 1)
of work-hours and y represents the available capital
per week. Find the marginal productivity of x and the
marginal productivity of y.
z ln ( y + 1)
=
.
Solution:The marginal productivity of x is
x
2x
z
x
.
The marginal productivity of y is =
y y + 1
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AppliedMaxima&Minima
Partial derivatives are used to solve optimization
problems involving non-linear functions. The
process of solving such problem includes finding
the relative maxima and minima for the function.
The procedure:
1. Find fx and fy.
2. Find the points the satisfy both fx = 0 and fy = 0.
These are the critical points.
3. Find all the second partial derivatives.
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AppliedMaxima&Minima
4. Evaluate D at each critical point.
D( x, y ) = f xx f yy ( f xy )
2
5. Use the test for maxima or minima to determine
whether relative maxima or minima occurs.
D(x, y)
fxx (x,y)
Interpretation
+
+
Relative minimum at (x, y).
+
-
Relative maximum at (x, y).
0
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Neither maximum or minimum at (x, y).
Test is inconclusive.
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Example
3
2
Test z = x + y + 6 xy + 24 x
minima.
Solution:
for maxima and
First partial derivatives,
z x = 3 x 2 + 6 y + 24
z y = 2 y + 6x
Solve for x and y when zx = 0 and zy
= 0, 3 x 2 + 6 y + 24 = 0 .........Equation(1)
2 y + 6x = 0
2 y = 6 x
y = 3 x
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.........Equation(2)
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Continued...
Substitute equation (2) in (1):
3 x + 2 6( 3x ) + 24 = 0
3 x 2 18 x + 24 = 0
x2 6x + 8 = 0
( x 4) ( x 2) = 0
x = 4 or 2
At x = 4, y = -3(4) = -12 and at x = 2, y = -3(2) = -6.
Find all second partial derivatives:
z xx = 6 x
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z xy = 6
14
Continued...
( ) = ( 6 x ) ( 2) ( 6)
Hence, D = z z z
xx yy
xy
2
2
= 12 x 36
D
At x = 4 and y = -12, = 12( 4 ) 36 = 12
At x = 2 and y = -6, D = 12( 2 ) 36 = 12
Since D > 0 and zxx > 0 at the critical point (4, -12), z has
a relative minimum point at x = 4 and y = -12.
For the point (2, -16), D is negative thus the test is
inconclusive.
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Problem
Suppose that x units of one input and y units of a
P = 40 x + 50 y x 2 y 2 xy
second input result in
units of product. Determine the inputs of x and y
that will maximize P. What is the maximum
pSolution: First partial derivatives,
roduction?
Px = 40 2 x y
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Continued...
Solve for x and y when Px = 0 and Py
= 40 2 x y = 0 .........Equation(1)
0,
50 2 y x = 0
x = 50 2 y
.........Equation(2)
Substitute equation (2) in (1):
40 2( 50 2 y ) y = 0
40 100 + 4 y y = 0
3 y = 60
y = 20
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x = 50 2( 20 ) = 10
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Continued...
Find all second partial derivatives:
Pxx = 2
Pyy = 2
Pxy = 1
Hence
2
2
D = Pxx Pyy Pxy = ( 2 ) ( 2 ) ( 1) = 4 1 = 3
,
Since D > 0 and Pxx < 0 at the critical point (10, 20), P
has a relative maximum point. Therefore, P is
maximized when 10 units of the first input and 20 units
of the second input are produced. The maximum
production is
2
2
()
P = 40(10 ) + 50( 20) 10 20 10( 20) = 700
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LagrangeMultipliers
We can obtain maxima and minima for a function
z = f(x, y) subject to a constraint g(x,y) = 0 by
using the method of Lagrange Multipliers.
For example, a consumer may want to maximize
utility derived from the consumption of
commodities subject to budget constraint or a
manufacturer may want to produce a box with a
fixed volume so that the least of amount of
material is used.
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MethodofLagrangeMultipliers
Consider a function z = f(x, y) subject to a single
constraint g(x, y) = 0.
We introduce a new variable, , called the
Lagrange multiplier and construct the new
objective function, ) = f ( x, y ) + g ( x, y )
F ( x, y ,
Then, find the critical points of F that will satisfy
Fx = 0, Fy = 0, F = 0
This critical points will satisfy the constraint g(x, y)
and will be the critical points for f(x, y).
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MethodofLagrangeMultipliers
However, this method will not tell us whether the
critical points correspond to maxima or minima,
but this can be determined by testing according to
a similar procedure that is used for unconstrained
maxima and minima for finding the relative
maxima and minima of F.
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Example
2
z
Find the maximum value of = x y
x 0, y 0.
subject + y = 9,
x to
Solution:
The objective function:F ( x, y, ) = x 2 y + ( x + y 9 )
First partial derivatives,
Fx = 2 xy + = 0 ..............Equation(1)
2
= x 2 ....Equation(2)
Fy = x + = 0
F = x + y 9 = 0 ..............Equation(3)
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Continued...
Substitute equation (2) in (1):
(
)
2 xy + x = 0
2
x 2 + 2 xy = 0
x( x 2 y ) = 0
x = 0 or x = 2 y
Since x = 0 could not make z = x2y a maximum,
substitute
x = 2y in equation (3):
( 2 y) + y 9 = 0
Hence x = 2( 3) = 6
,
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Continued...
To determine whether the critical points maximizes F,
Fyy = 0
Fxx = 2 y
Fxy = 2 x
Hence D = F F ( F ) 2 = ( 6 ) ( 0 ) (12) 2 = 144
xx yy
xy
,
Since D < 0, the test inconclusive. Test for several pair
of points that satisfy the constraint x + y 9 =0 and see
which point gives a maximum value to z.
z = ( 52 )( 4 ) = 100
z = ( 6 2 )( 3) = 108
z = ( 7 2 )( 2 ) = 98
Therefore, the critical point x = 6 and y = 3 gives a
maximum value to z = 108 while satisfying the
constraint x + y = 9.
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LearningOutcomes
Able to solve unconstrained optimization problem.
Able to solve a single constrained optimization
problem.
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