(chapter 7) Partial Derivates

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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

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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.

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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)

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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 =

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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

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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 ??.

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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 ??

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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.

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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??.

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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

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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

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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)

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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).

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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.

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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.

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