Lecture Note #14 Chapter 7-(2)

Business

Mathematics

1

Applications of Partial

Differentiation

1. Unconstrained Optimization

2. Constrained Opimization

2

Optimization Optimization means either maximization or minimization.

Optimization is finding maximum or minimum points which

are called optimum points

Optimization

Maximization

Minimization

Locating maximum point

Locating minimum point

Optimization Locating

Optimum point

Optimum point Maximum point Minimum point

3

Unconstrained Optimization

Optimization for functions of one variable: y=f(x)

The three steps for finding optimum points for y=f(x)

Step1: Find the first and second derivatives: 𝑑𝑦

𝑑𝑥 𝑎𝑛𝑑

𝑑2𝑦

𝑑𝑥2

Step2: Equate the first derivative to zero, and solve this equation to find x-coordinate of the potential turning point(s).

Step3: Use second derivative to determine the nature of the

turning point(s). Evaluate 𝑑2𝑦

𝑑𝑥2 at each turning point.

If 𝑑2𝑦

𝑑𝑥2 at turning point > 0, the point is a minimum

If 𝑑2𝑦

𝑑𝑥2 at turning point < 0, the point is a maximum

If 𝑑2𝑦

𝑑𝑥2 at turning point = 0, the point may be a PoI

4

Unconstrained Optimization

Optimization for functions of two variable: z=f(x,y)

The methods for finding optimum points or PoI for

functions of two variables are simply extensions of

those used for functions of one variable.

The method consists of three steps as for functions of

one variable. This will be explained on the next page.

5

Unconstrained Optimization

Optimization for functions of two variable: z=f(x,y)

The three steps for finding optimum points for z=f(x,y)

Step1: Find the first and second derivatives:

𝜕𝑧

𝜕𝑥

𝜕𝑧

𝜕𝑦

𝜕2𝑧

𝜕𝑥2

𝜕2𝑧

𝜕𝑦2

𝜕2𝑧

𝜕𝑥𝜕𝑦

Step2: Equate the first derivative to zero, and solve

these equations to find x- and y-coordinates of the

potential turning point(s).

𝜕𝑧

𝜕𝑥= 0,

𝜕𝑧

𝜕𝑦= 0

6

Unconstrained Optimization

Optimization for functions of two variable: z=f(x,y)

The three steps for finding optimum points for z=f(x,y)

Step3: Use the second derivatives to determine the nature of the

turning point(s). Evaluate all second derivatives at x- and y-

coordinates of the turning points.

Here, need to define ∆; ∆= 𝜕2𝑧

𝜕𝑥2∙

𝜕2𝑧

𝜕𝑦2 − (

𝜕2𝑧

𝜕𝑥𝜕𝑦)

2

If 𝑑2𝑧

𝑑𝑥2 > 0 and

𝑑2𝑧

𝑑𝑦2 > 0 and ∆ > 0, the point is a minimum

If 𝑑2𝑧

𝑑𝑥2 < 0 and

𝑑2𝑧

𝑑𝑦2 < 0 and ∆ > 0, the point is a maximum

If 𝑑2𝑦

𝑑𝑥2

𝑑2𝑧

𝑑𝑦2 > 0 and ∆ < 0 , the point is a point of inflection

If 𝑑2𝑦

𝑑𝑥2

𝑑2𝑧

𝑑𝑦2 < 0 and ∆ < 0 , the point is a saddle point

7

Unconstrained Optimization Optimization for functions of two variable: z=f(x,y) The three steps for finding optimum points for z=f(x,y)

Step3(continued): Use the second derivatives to determine the nature of the turning point(s). Evaluate all second derivatives at x- and y-coordinates of the turning points.

If ∆= 0 then there is no conclusion.

A saddle point is a point which is a maximum when viewed along one axis but is a minimum when viewed along the other axis. (See Figure 7.9 (c))

Note: It is possible to classify all given functions in this text according to the second-order conditions outlined above. But for those functions for which these conditions fail, then higher-order derivatives are required, but this is beyond the scope of this text.

8

9

Unconstrained Optimization

Worked Example 7.14

Given that z = f(x,y) = 𝑥2 + 𝑦2 + 20

Find the optimum point and determine the nature of the optimum point.

Step1: Find the first and second derivatives:

𝑧𝑥 = 2𝑥, 𝑧𝑦 = 2𝑦, 𝑧𝑥𝑥 = 2, 𝑧𝑦𝑦 = 2, 𝑧𝑥𝑦 = 0

Step2: Equate the first derivatives to zero, and solve for x- and y-coordinates of the turning point:

𝑧𝑥 = 2𝑥 = 0, 𝑧𝑦 = 2𝑦 = 0 𝑥 = 0, 𝑦 = 0

Step3: Use the second derivatives to determine the nature of the turning point: 𝑧𝑥𝑥 = 2 > 0, 𝑧𝑦𝑦 = 2 > 0

and ∆= 𝑧𝑥𝑥 ∙ 𝑧𝑦𝑦 − (𝑧𝑥𝑦)2= 2 ∙ 2 − 02 = 4 > 0

Therefore the turning point is a minimum. The minimum point is at x=0, y=0 and the minimum value of z is 20.

10

Unconstrained Optimization Worked Example 7.16: To Maximize Profit for a Multi-Product Firm in a Perfectly Competitive Market A perfectly competitive firm produces two goods, X and Y, which are sold at $54 and $52 per unit, respectively. The firm’s TC function is given by

TC = f(x,y) = 3𝑥2 + 3𝑥𝑦 + 2𝑦2 − 100 Find the quantities of each good which must be produced and sold to maximize the profit. What is the maximum profit?

Since a perfectly competitive firm charges the same price regardless of the number of units sold, the Total Revenue 𝑇𝑅 = 𝑥 ∙ 𝑃𝑋 + 𝑦 ∙ 𝑃𝑌 = 54𝑥 + 52𝑦 And Profit 𝜋 = 𝑇𝑅 − 𝑇𝐶 = 54𝑥 + 52𝑦 − 3𝑥2 − 3𝑥𝑦 − 2𝑦2 + 100

Step1: Find the first- and second-order partial derivatives:

𝜋𝑥 = 54 − 6𝑥 − 3𝑦, 𝜋𝑦 = 52 − 3𝑥 − 4𝑦, 𝜋𝑥𝑥 = −6, 𝜋𝑦𝑦 = −4, 𝜋𝑥𝑦 = −3

Step2: Equate the first-order partial derivatives to zero, and solve for x- and y-coordinates of the turning point:

𝜋𝑥 = 54 − 6𝑥 − 3𝑦 = 0, 𝜋𝑦 = 52 − 3𝑥 − 4𝑦 = 0 𝒙 = 𝟒, 𝒚 = 𝟏𝟎 Step3: Use the second derivatives to determine the nature of the turning point:

𝜋𝑥𝑥 = −6 < 0, 𝜋𝑦𝑦 = −4 < 0

and ∆= 𝜋𝑥𝑥 ∙ 𝜋𝑦𝑦 − (𝜋𝑥𝑦)2= (-6)(-4)−(−3)2= 15 > 0

when x=4 and y=10 𝜋 = 54 ∙ 4 + 52 ∙ 10 − 3 ∙ 42-3∙ 4 ∙ 10 − 2 ∙ 102 + 100 = 468 Therefore the turning point is a maximum. The maximum point is at x=4, y=10 and the maximum profit is $468.

11

Unconstrained Optimization

Price Discrimination

A monopolist may charge different prices in different markets for

the same product. This is called Price Discrimination.

Consider a monopolist producing a single product which is sold in

two separate markets, market 1 and market 2, each with a different

demand function.

What would be the prices which should be charged in the two

markets to maximize profit?

Next page

12

Unconstrained Optimization

Price Discrimination

Start by writing down the equation for the profit:

𝜋 = 𝑇𝑅 − 𝑇𝐶

𝑇𝑅 = 𝑇𝑅1 + 𝑇𝑅2 = 𝑃1𝑥1 + 𝑃2𝑥2

So, 𝜋 = 𝑇𝑅 − 𝑇𝐶 = 𝑇𝑅1 + 𝑇𝑅2 − 𝑇𝐶 = 𝑃1𝑥1 + 𝑃2𝑥2 − TC

where 𝑇𝑅1=TR from market 1

𝑇𝑅2=TR from market 2

𝑃1= price of the product in market 1

𝑃2= price of the product in market 2

𝑥1= quantity sold in market 1

𝑥2= quantity sold in market 2

13

Unconstrained Optimization

Price Discrimination

Step1: Find first- and second –order partial derivatives:

𝜕𝜋

𝜕𝑥1=

𝜕𝑇𝑅1

𝜕𝑥1−

𝜕𝑇𝐶

𝜕𝑥1 𝜋𝑥1 = 𝑇𝑅𝑥1 − 𝑇𝐶𝑥1 = 𝑀𝑅1 − 𝑀𝐶1

𝜕𝜋

𝜕𝑥2=

𝜕𝑇𝑅2

𝜕𝑥2−

𝜕𝑇𝐶

𝜕𝑥2 𝜋𝑥2 = 𝑇𝑅𝑥2 − 𝑇𝐶𝑥2 = 𝑀𝑅2 − 𝑀𝐶2

𝜋𝑥1𝑥1 = 𝑇𝑅𝑥1𝑥1 − 𝑇𝐶𝑥1𝑥1

𝜋𝑥2𝑥2= 𝑇𝑅𝑥2𝑥2 − 𝑇𝐶𝑥2𝑥2

𝜋𝑥1𝑥2 = 𝑇𝑅𝑥1𝑥2 − 𝑇𝐶𝑥1𝑥2

14

Unconstrained Optimization Price Discrimination Step2: Equate the first-order partial derivatives to zero and solve:

𝜋𝑥1= 0 𝑇𝑅𝑥1 − 𝑇𝐶𝑥1 = 𝑀𝑅1 − 𝑀𝐶1 = 0 𝑀𝑅1 = 𝑀𝐶1 𝜋𝑥2 = 0 𝑇𝑅𝑥2 − 𝑇𝐶𝑥2 = 𝑀𝑅2 − 𝑀𝐶2 = 0 𝑀𝑅2 = 𝑀𝐶2

In other words, MR=MC for each market for maximum profit.

Since the single product is produced by one firm, the cost function TC is the same for each market.

Consequently, the MC is the same for each market.

Therefore, 𝑀𝐶1 = 𝑀𝐶2 = 𝑀𝐶 Thus the first-order condition for maximum profit is

𝑀𝐶1 = 𝑀𝐶2 = 𝑀𝐶

Step3: The second-order conditions must be used to classify the point as a maximum, minimum or otherwise. The method is the same as explained before, so is omitted.

15

Unconstrained Optimization Worked Example 7.17: Monopolist: Price and Non-Price Discrimination Demand functions: 𝑃1 = 50 − 4𝑥1 and 𝑃2 = 80 − 3𝑥2 Total Cost function: 𝑇𝐶 = 120 + 8𝑥 where 𝑥 = 𝑥1 + 𝑥2

(a) Find the price and quantity of the good in each market which maximizes the profit.

𝑇𝑅1 = 𝑃1𝑥1 = (50 − 4𝑥1)𝑥1 = 50𝑥1 − 4𝑥12

𝑇𝑅2= 𝑃2𝑥2 = (80 − 3𝑥2)𝑥2 = 80𝑥2 − 3𝑥22

Then, 𝜋 = 𝑇𝑅1 + 𝑇𝑅2 − 𝑇𝐶 = 50𝑥1 − 4𝑥1

2 + 80𝑥2 − 3𝑥22 − 120 − 8(𝑥1 + 𝑥2)

= 42𝑥1 − 4𝑥12 + 72𝑥2 − 3𝑥2

2 − 120

16

Unconstrained Optimization

Price Discrimination Step1: Find first- and second –order partial derivatives:

𝜋𝑥1 =𝜕𝜋

𝜕𝑥1= 42 − 8𝑥1 (= 𝑀𝑅1 − 𝑀𝐶1)

𝜋𝑥2 =𝜕𝜋

𝜕𝑥2= 72 − 6𝑥2 (= 𝑀𝑅2 − 𝑀𝐶2)

𝜋𝑥1𝑥1 = −8 𝜋𝑥2𝑥2= −6 𝜋𝑥1𝑥2 =0

Step2: Equate the first-order partial derivatives to zero and solve:

𝜋𝑥1= 0 42 − 8𝑥1 = 0 𝐱𝟏 =5.25

𝜋𝑥2 = 0 72 − 6𝑥2 = 0 𝐱𝟐 =12

And when 𝐱𝟏 =5.25 and 𝐱𝟐 =12

𝜋 = 42𝑥1 − 4𝑥12 + 72𝑥2 − 3𝑥2

2 − 120

= 42 ∙ 5.25 − 4 ∙ 5.252 + 72 ∙ 12 − 3 ∙ 122 − 120 = 𝟒𝟐𝟐. 𝟐𝟓

17

Unconstrained Optimization

Price Discrimination

Step3: Use the second derivatives to determine the nature of the

turning point:

𝜋𝑥1𝑥1 = −8 < 0 𝜋𝑥2𝑥2= −6 < 0

and ∆= 𝜋𝑥1𝑥1 ∙ 𝜋𝑥2𝑥2 −(𝜋𝑥1𝑥2)2= (-8)(-6)−(0)2= 48 > 0

So, when 𝒙𝟏 =5.25 and 𝒙𝟐 =12 the profit is the maximum. The maximum profit at 𝑥1 =5.25 and 𝑥2 =12 is 422.25.

The price in each market for the product is

𝑃1 = 50 − 4𝑥1 = 50 − 4 ∙ 5.25 = 𝟐𝟗

and 𝑃2 = 80 − 3𝑥2 = 80 − 3 ∙ 12 = 𝟒𝟒

18

Unconstrained Optimization Worked Example 7.17:

Monopolist: Price and Non-Price Discrimination Demand functions: 𝑃1 = 50 − 4𝑥1 and 𝑃2 = 80 − 3𝑥2

Total Cost function: 𝑇𝐶 = 120 + 8𝑥 where 𝑥 = 𝑥1 + 𝑥2

(b) Determine the price elasticity of demand for each market when the profit is maximized.

When profit is maximized x1 =5.25 , P1 =29 and x2 =12 , P2 =44.

Substitute these values into the formula for elasticity:

𝜀1 =𝑑𝑥1

𝑑𝑃1

𝑃1

𝑥1 =

1𝑑𝑃1𝑑𝑥1

𝑃1

𝑥1 =

1

−4∙

29

5.25= −1.381

𝜀2 =𝑑𝑥2

𝑑𝑃2

𝑃2

𝑥2 =

1𝑑𝑃2𝑑𝑥2

𝑃2

𝑥2 =

1

−3∙

44

12= −1.222

Since |𝜀1| > 1 and |𝜀2| > 1 the demands of both markets are elastic.

But |𝜀1| > |𝜀2| tells us that market 2 is relatively less elastic. Therefore, the higher price (P2 =44 > P1 =29) is charged in the less elastic market which is market 2.

19

Unconstrained Optimization

Worked Example 7.17:

Monopolist: Price and Non-Price Discrimination Demand functions: 𝑃1 = 50 − 4𝑥1 and 𝑃2 = 80 − 3𝑥2

Total Cost function: 𝑇𝐶 = 120 + 8𝑥 where 𝑥 = 𝑥1 + 𝑥2

(c) Find the price and quantity of the good in each market which

maximizes profit when the monopolist does not use price

discrimination, that is, 𝑃1 = 𝑃2.

Since 𝑃1 = 𝑃2, let 𝑃1 = 𝑃2 = 𝑃.

Then 50 − 4𝑥1 = 𝑃 ⋯ (1) and 80 − 3𝑥2 = 𝑃 ⋯ (2)

(1)× 3 3(50 − 4𝑥1) = 3𝑃 ⋯ (3)

(2)× 4 4(80−3𝑥2) = 4𝑃 ⋯ (4)

(3)+(4) 470-12(𝑥1 + 𝑥2) = 7P 7𝑃 = 470 − 12𝑥 (where 𝑥 = 𝑥1 + 𝑥2) 𝑃 = 67.14 − 1.714𝑥

20

Unconstrained Optimization

Worked Example 7.17:

Monopolist: Price and Non-Price Discrimination

(c) (continued)

Now we have the overall demand function 𝑃 = 67.14 − 1.714𝑥

This becomes the optimization for function of one variable.

The profit function is 𝜋 = 𝑇𝑅 − 𝑇𝐶 = 𝑃 ∙ 𝑥 −(120 + 8𝑥)

= (67.14−1.714𝑥)𝑥 −(120 + 8𝑥)

= −1.714𝑥2 + 59.14𝑥 − 120

Equate 𝜋′ to zero: 𝜋′ =𝑑𝜋

𝑑𝑥= 𝑀𝑅 − 𝑀𝐶 = −3.428𝑥 + 59.14 = 0

𝑥 = 17.252

21

Unconstrained Optimization

Worked Example 7.17:

Monopolist: Price and Non-Price Discrimination (c) (continued)

Evaluate 𝜋′′ at 𝑥 = 17.252

𝜋′′ =𝑑2𝜋

𝑑𝑥2= 𝑀𝑅′ − 𝑀𝐶′ = −3.428 < 0

The profit when 𝑥 = 17.252 is

𝜋 = −1.714𝑥2 + 59.14𝑥 − 120 = 390.1

Therefore, the profit is maximized when 𝑥 = 17.252 and the maximum profit is 390.1

(The price to be charged is 𝑃 = 67.14 − 1.714𝑥 = 67.14 − 1.714 ∙17.252 = 37.57)

22

Unconstrained Optimization

Worked Example 7.17:

Monopolist: Comparison between Price Discrimination and Non-Price Discrimination

When price discrimination is not used(or not allowed) the common price is determined at some level between 𝑃1 and 𝑃2, and the profit reduces.

(Also, note that less elastic market is charged higher price.)

23

Discrimination Non-discrimination

Price 𝑃1 = 29, 𝑃2 = 44 𝑃 = 37.57

Quantity sold 𝑥1 = 5.25, 𝑥2 = 12 𝑥 = 𝑥1 + 𝑥2 = 17.25

Profit 𝜋 = 422.25 𝜋 = 390.1