QQM1023 Managerial Mathematics
f x ( x, y ) =
8.1 First and Second Order Partial Derivatives 8.1.1 First Order Partial Derivatives Let z = f ( x, y ) is a function of two variables. We write: fx (x, y),
∂z ∂f , or for the first order partial derivative of z = f(x, y) with ∂x ∂x
respect to x fy (x, y),
∂z ∂f for the first order partial derivative of z = f(x, y) with , or ∂y ∂y
respect to y
Example 1: Given f(x, y) = 2x + 3y. Find fx and fy. Solution : To find fx (x, y) treat y as a constant and x as a variable.
This operation is called as a partial derivative of f with respect to x. To find fy(x, y), treat y as a variable and x as a constant.
This operation is called partial derivative of
Chapter 8: Partial Derivatives
f with respect to y. 204
δf δx
QQM1023 Managerial Mathematics
Example 2: Let f(x, y) = 8x2 + 2y. Find (i)
fx and fy.
(ii) f(1, 1), fx(2, 1) and fy(1, 3).
Example 3: Let f(x, y) = 3x + bxy. Find the value of b if f(1, 3) = 9. Then, find fx and fy.
Chapter 8: Partial Derivatives
205
QQM1023 Managerial Mathematics
EXERCISE : Find fx and fy . Then find fx(2, –1) and fy(– 4, 3) for the following functions.
a)
f(x, y) = – 2xy + 6y3 + 2
b)
f(x, y) = 3x3y2
c)
f(x, y) = ex + y
d)
f(x, y) = – 5e3x – 4y
x 2 + y3 e) f ( x, y ) = x3 − y 2
f)
f(x, y) = ln (1 + 3x2y3)
Chapter 8: Partial Derivatives
206
QQM1023 Managerial Mathematics 8.1.2 Second-Order Partial Derivatives
For a function z = f(x, y), if the indicated partial derivative exist, then there are four second-order partial derivatives as given below.
∂ ∂z ∂ 2 z = = f xx ( x, y) ∂x ∂x ∂x 2
∂ ∂z ∂ 2 z = = f yy ( x, y) ∂y ∂y ∂y 2
∂ ∂z ∂ 2 z = = f xy ( x, y ) ∂y ∂x ∂y∂x
∂ ∂z ∂ 2 z = = f yx ( x, y) ∂x ∂y ∂x∂y
Example 4: Let f(x, y) = x2 + xy + y2. Find fx , fxx , fy , fyy , fxy and fyx Solution : fx = fxx meaning differentiate fx with respect to x
Æ fxx = fxy meaning differentiate fx with respect to y.
Æ fxy = Æ fy = fyy meaning differentiate fy with respect to y.
Æ fyy = fyx meaning differentiate fy with respect to x.
Obviously that fxy = fyx
Æ fyx =
Chapter 8: Partial Derivatives
207
QQM1023 Managerial Mathematics
Example 5: Let z = f(x, y) = x3 + 5y4. Find
∂2z ∂2z ∂2z ∂2z , , and . 2 2 ∂ x ∂ y ∂ y ∂ x ∂x ∂y
Solution:
∂z = ∂x
∂2z = 2 ∂x
∂2z = ∂y∂x
∂z = ∂y
∂2z = 2 ∂y
∂2z = ∂x∂y
EXERCISE 1. Find fx, fy, fxx, fyy dan fxy for the given functions. a) f(x, y) = 31xy2 + 2x5 – 7y3
b) f(x, y) = 13x3y4 – 6x + 11y2
c) f(x, y) = x8y2 + 9y2 – 6xy
d) f(x, y) = 7xy – 4x3 + 2xy2
Chapter 8: Partial Derivatives
208
QQM1023 Managerial Mathematics 8.2 Applications : Critical Points (maximum, minimum, saddle point) Finding Critical Points
Given f(x, y), we will find the critical point (a, b) when fx = 0 and fy = 0.
Example 6: Given f(x, y) = x2 – 4x + 2y2 + 4y + 7 Solution : Step 1
Find fx and fy
fx = fy = Step 2
Let fx and fy with 0 that is
fx = 0 =
---------------(1)
and fy = 0
---------------(2)
Thus, we get two equations.
Step 3
Determine the critical point by solving equations simultaneously.
Therefore, the critical point is ??.
Chapter 8: Partial Derivatives
209
QQM1023 Managerial Mathematics
Example 7: 2
2
Given f(x, y) = 2x + y – 2xy + 5x – 3y + 1. Determine the critical point. Solution :
S1
Find fx and fy
S2
Set fx = 0 and fy = 0
S3
Solve the equations (1) and (2)
Therefore, the critical point is ??
Chapter 8: Partial Derivatives
210
QQM1023 Managerial Mathematics Second-Derivative : Test to determine whether the critical point is maximum, minimum, or saddle point. Suppose
z = f(x, y) has continuous partial derivatives fxx, fyy, and fxy at all
points (x, y) near the critical point (a, b). Let D be the function defined by
D(a,b) = fxx(a, b) fyy(a, b) – [fxy(a, b)]2 Then 1.
If D(a,b) > 0, fxx(a, b) < 0, f has relative maximum at (a, b).
2.
If D(a,b) > 0, fxx(a, b) > 0, f has relative minimum at point (a, b).
3.
If D(a,b) < 0, there is neither relative maximum nor relative minimum at (a, b) and we called it as a saddle point.
4.
If D(a,b) = 0, no conclusion about an extremum at (a, b) can be drawn, and further analysis is required.
saddle point
Steps to determine whether the critical point is maximum, minimum, or saddle point. 1. Find all the critical points. 2. Find second order partial derivatives, fxx, fyy and fxy at the all critical points. 3. Apply the second-derivative test.
Chapter 8: Partial Derivatives
211
QQM1023 Managerial Mathematics
Example 8: From Example 1, testify whether the critical point is maximum, minimum or saddle point. Solution : From Example 1 f(x, y) = x2 – 4x + 2y2 + 4y + 7. We already solved Step 1 in Example 1. Step 1
Find the critical point
fx = 2x – 4 ; fy = 4y + 4 and the critical point is (2, – 1).
Step 2
Find fxx , fyy and fxy at (2, –1)
fxx = Thus
fxx(2,–1) =
Step 3
fyy = ,
and
fxy =
fyy(2,–1) =
and
fxy(2, –1) =
Apply second-derivative test
D(2, –1) = [fxx(2, –1)][fyy(2,–1)] – [fxy(2,–1)]2 = = Therefore (2, –1) is max/min/saddle point??.
Chapter 8: Partial Derivatives
212
QQM1023 Managerial Mathematics
EXERCISE: Find the critical points of the functions. For each critical point, determine, by the second-derivative test, whether it corresponds to a maximum, minimum, or saddle point. (a)
f(x, y) = x2 + 3y2 + 4x– 9y + 3
(b)
f(x, y) = y – y2 – 3x – 6x2
(c)
f(x, y) = x3 – 3xy + y2 + y – 5
(d)
f(l, k) = 2lk – l2 + 264k – 10l – 2k2
Chapter 8: Partial Derivatives
213
QQM1023 Managerial Mathematics 8.3 Applications : Lagrange Multiplier
Suppose we have an objective function f(x, y) subject to the constraint g(x, y)=0. We construct a new function F of three variables x, y, λ called the Lagrange function defined by the following:
F (x, y, λ) = f(x, y) + λg(x, y). = objective + λ(constraint)
It can be shown that if (a, b) is a critical point of f, subject to the constraint g(x, y)=0, there exists a value of λ, say λ0, such that (a, b, λ0) is a critical point of F. The number λ0 is called a Lagrange multiplier. Thus, to find critical points of f, subject to the constrain g(x, y), we instead find critical points of F. These are obtained by solving the equations
Fx(x, y, λ) = 0 Fy(x, y λ) = 0 Fλ (x, y, λ) = 0
Chapter 8: Partial Derivatives
214
QQM 1023 Managerial Mathematics
Example 9: Find the critical point of f(x, y) = 5x2 + 6y2 – xy subject to the constraint x + 2y = 24. Solution :
Step 1
Rewrite the constraint in the form g(x, y) = 0.
The constraint becomes x + 2y – 24 = 0,
Step 2
with
g(x, y) = x + 2y – 24.
Form the Lagrange function F(x, y, λ).
Here F(x, y, λ) = f(x, y) + λ g(x, y) = 5x2 + 6y2 – xy + λ (x + 2y – 24) = 5x2 + 6y2 – xy + λ x + 2λy – 24λ
Step 3
Find Fx, Fy, and Fλ. Fx = 10x – y + λ Fy = 12y – x + 2λ Fλ = x + 2y – 24
Step 4
Form the system of equations Fx = 0, Fy = 0 and Fλ = 0. 10x – y + λ = 0 ------------(1) 12y – x + 2λ = 0 ------------(2) x + 2y – 24 = 0 ------------(3)
Chapter 8: Partial Derivatives
215
QQM 1023 Managerial Mathematics Step 5
Solve the system of equations from step 4 for x, y and λ.
One way to solve this system is to begin by solving each of the first two equations for λ, then set the two results equal and simplify, as follows. 10x – y + λ = 0
becomes
λ = – 10x + y
12y – x + 2λ = 0
becomes
λ=
– 10x + y =
x −12 y 2
x −12 y 2
Equalize the λ.
–20x + 2y = x – 12y –21x = –14y x=
2y 3
Now, substitute
2y 3
for x in equation (3). x + 2y – 24 = 0
2y 3
+ 2y – 24 = 0
2y + 6y – 72 = 0
Let x = 2 y . 3
8y = 72 y=9
Since x =
2y 3
and y = 9, x = 6. It is not necessary to find the value of λ.
Therefore, the critical point of f(x, y) = 5x2 + 6y2 – xy subject to the constraint x + 2y = 24 is (6, 9).
Chapter 8: Partial Derivatives
216
QQM 1023 Managerial Mathematics
EXERCISE: 1.
Find the critical points of f(x, y) = x2 + 4y2 + 6 subject to 2x – 8y = 20.
2.
Find the critical points of f(x, y) = 3x2 + 4y2 – xy – 2 subject to 2x + y = 21.
Chapter 8: Partial Derivatives
217
QQM 1023 Managerial Mathematics 3.
Maximizing Output : A factory produce two types of bolt namely type KK and type LL. The total number of bolts to be produced is determined by the function f(l, k) = 12l + 20k – l2 – 2k2, where k is the number of KK bolt to be produced and LL is the number of LL bolt to be produced. The cost to produce one unit of KK bolt is RM4 and for LL bolt is RM8 per unit, respectively. If the firm wants the total cost to produce these two types of bolts to be RM88, find the greatest output possible, subject to this budget constraint.
Chapter 8: Partial Derivatives
218