Calculus Applied to Business Part04

8/21/2019 Calculus Applied to Business Part04

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Calculus Applied to Business and Economics

Rules in Finding the Derivative of a Function

Prof. Kenneth James T. Nuguid

May 3, 2013

Prof. Kenneth James T. Nuguid Calculus Applied to Business and Economics


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Average Rate of Change


The average rate of change of y with respect to x , as x changesfrom x 1 to x 2 is the ratio of the change in output to the change ininput:

Average Rate of Change = y 2 − y 1x 2 − x 1

= f (x 2) − f (x 1)

x 2 − x 1.



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Average Rate of Production

Example 1. The following graph shows the total production of suits by Raggs, Ltd., during one morning of work. What was thenumber of suits produced at Raggs, Ltd., from 9 A.M. to 11 A.M.?



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Solution: At 9 A.M., 20 suits had been produced. At 11 A.M., 64suits had been produced. From 9 A.M. to 11 A.M., the average

number of suits produced was64suits − 20suits

3 − 1hour = 44suits

2hr = 22

suits

hr .



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Difference Quotient

If we change x 1 by x and x 2 by x + h in the definition of averagerate of change, we get

Average Rate of Change = f (x + h) − f (x )

(x + h) − x = f (x + h) − f (x )

h .

The average rate of change is also called the difference quotient.



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Economic Utility

Exercise 1a. Utility is a type of function that occurs in economics.

When a consumer receives x units of a certain product, a certainamount of pleasure, or utility U , is derived. The following is agraph of a typical utility function.



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Economic Utility

a. Find the average rate of change of U as x changes from 0 to 1;from 1 to 2; from 2 to 3; from 3 to 4.

b. Why do you think the average rates of change are decreasing asx increases?

Compound Interest

Exercise 1b. The amount of money, in a savings account that pays6% interest, compounded quarterly for t years, when an initial

investment of P20,000 is made, is given byA(t ) = 20, 000(1.015)4t .

Find A(5)−A(3)5−3 , and interpret this result.



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Instantaneous Rate of Change

Instantaneous Rate of Change

The slope of the tangent line at (x , f (x )) is

m = limh→0

f (x + h) − f (x )h

.

This limit is also called the instantaneous rate of change.


f


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Derivative of a Function


For a function y = f (x ) its derivative is the function f (x ) definedby

f (x ) = limh→0

f (x + h) − f (x )h

,

provided that the limit exists. If f (x ) exists then we say that f is

differentiable at x .

Example 2. For f (x ) = x 2, find f (x ). Find f (−3) and f (4).Solution:

f

(x ) = limh→0

f (x + h)−

f (x )

h = limh→0

(x + h)2

−x 2

h

= limh→0

x 2 + 2xh + h2 − x 2h

= limh→0

h(2x + h)

h = lim

h→02x + h = 2x .

Thus, f (−

3) = 2(−

3) =−

6 and f (4) = 2(4) = 8.


D i i f F i


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Exercise 2a. For f (x ) = x 3. Find f (x ). Find f (−1) and f (1.5).At which values of x is the derivative 0?Exercise 2b. For f (x ) = 1

x . Find f (x ). Then find f (−1) and f (2)

At which values of x is the derivative 0?.

A function is not differentiable at a point x = a if:

a. there is a discontinuity at x = a.

b. there is a corner at x = a or

c. there is a vertical tangent at x = a.


L ib i N i


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Leibniz Notation

Leibniz Notation

Let y be a function of x , i.e. y = f (x ), then the derivative of y

with respect to x or the derivative of the function f (x ) withrespect to x is given by

dy

dx = y = f (x ).

When we wish to evaluate a derivative at a number, e.g. at x = 2,we write

dy

dx

x =2

= f (2).

Example. If y = f (x ) = x 2 then dy dx

= y = f (x ) = 2x and

dy

dx x =2 = f (2) = 2(2) = 4.


B i R l f Diff ti ti


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Basic Rules of Differentiation

Differentiation Rules

i. For any constant c , if y = c , then d dx

c = 0.

ii. For any real number n, if y = x n, then d dx

x n = n · x n−1.iii. For any constant c , d

dx [c · f (x )] = c · d

dx f (x ).

iv. The derivatie of a sum is the sum of the derivatives:

d

dx [f (x ) ± g (x )] = d

dx f (x ) ± d

dx g (x ).

v. The derivatie of a sum is given by:

d

dx [f (x ) · g (x )] = d

dx f (x ) · [g (x )] + d

dx g (x ) · [f (x )].


B i R l f Diff ti ti


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Example 3. Find each of the following derivatives:

i. d dx

7x 4

ii. d dx

( 15x 2

)

iii. d dx

(5x 3

−7)

iv. d dx

24 −√ x + 5

x

Solution:

i. d dx

7x 4 = 7 d dx

x 4 = 7(4x 4−1) = 28x 3

ii. d dx

( 15x 2

) = d dx

(5x −2) = 5((

−2)x −2−1) =

−10x −3

iii. d dx (5x 3 − 7) = 5( d dx x 3) − d dx 7 = 5(3x 2) − 0 = 15x 2iv.

d

dx

24 −√ x + 5

x

= 0 − d

dx √

x

+ 5

d

dx

1

x

= − d dx

x 1/2

+ 5 d

dx

x −1

= −1

2x

12−1 + 5(−1)(x −1−1

=−

1

2

x −12

−5(x −2) =

−1

2√ x −5

x 2 .


B si R l s f Diff ti ti


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Exercise 3a. Find each of the following derivatives:

i. d dx

4√

x + 6x 4

ii. d dx x 0.7 + 3x

4 − 0.01x 2iii. d

dx ( 2

3x 4)

iv. d dx

(−x 3 + 6x 2)v. d

dx 3x 5 + 2 3√ x + 1

3x 2 +

√ 5

Exercise 3b. Evaluate the following:

i. If f (x ) = x 2 + 4x − 5 find f (10).

ii. If y = x 3 + +2x − 5, find dy dx

x =−

2

.

iii. If y = x + 2x 3

, find dy dx

x =1

.

iv. If y = 3√

x +√

x , find dy dx x

=64

.




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Example 4. Find d dx

(x 4 − 2x 3 − 7)(3x 2 − 5x ) without simplifying.

Solution: Let f (x ) = x 4 − 2x 3 − 7 and g (x ) = 3x 2 − 5x thendifferentiating f (x ) and g (x ) we get

f (x ) = 4x 3 − 6x 2g (x ) = 6x − 5.

Hence, applying the product rule, we obtaind

dx [f (x ) · g (x ) = f (x ) · g (x ) + g (x ) · f (x )

d

dx (x 4 − 2x 3 − 7)(3x 2 − 5x ) = (x

4 − 2x 3 − 7)(6x − 5)+ (3x 2 − 5x )(4x 3 − 6x 2).

Exercise 4. Use the product rule to differentiate

y = (2x 5 + x

−1)(3x

−2).




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Example 5. Find d dx

x 5

x 2

without simplifying.

Solution: Let f (x ) = x 5 and g (x ) = x 2 then differentiating f (x )

and g (x ) we get

f (x ) = 5x 4

g (x ) = 2x .

Hence, applying the quotient rule, we obtain

d

dx

f (x )

g (x )

=

g (x ) · f (x ) − f (x ) · g (x )[g (x )]2

d

dx x 5

x 2

=

x 2

·5x 4

−x 5

·2x

(x 2)2

= 5x 6 − 2x 6

x 4 =

3x 6

x 4 = 3x 2.

Exercise 5. Use the quotient rule to differentiate y = 1+x 2

x 3 .


Applications


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Applications

Instantaneous Rate of ChangeExample 6. Basketball Ticket Prices . The average price, in pesos,of a ticket for a PBA basketball game x years after 1990 can beestimated by

p (x ) = 94.1−

1.9x + 0.9x 2.

a. Find the rate of change of the average ticket price with respectto the year, dp

dx .

b. What is the average ticket price in 2010?c. What is the rate of change of the average ticket price in 2010?


Applications


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Applications

Solution:

a. dp dx

= p (x ) = −1.9 + 0.9(2)x = −1.9 + 1.8x .b. The average ticket price x yrs after 1990 is given by p (x ) and

2010 is 20 years after 1990. Hence, the average ticket price in2010 is given by

p (20) = 9.41 − 1.9(20) + 0.9(202) = P 331.41.

c. The rate of change of the average ticket price in 2010 is thevalue of p (x ) when x = 20. Hence,

p (20) = −1.9 + 1.8(20) = P 34.1 per year.


Applications


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Applications

Exercise 6a. Advertising . A firm estimates that it will sell N unitsof a product after spending a thousands of pesos on advertising,

where N (a) = −a2

+ 300a + 6.a. Find the rate of change of the number of units sold with respect

to the amount spent on advertising, dN da

= N (a).

b. How many units will be sold after spending P100,000?

c. What is the rate of change of a = 100?

Exercise 6b. Demand . A demand function for a certain product isgiven by D (p ) = 100 −√ p .a. Find the rate of change of quantity demanded with respect to

price, dD dp

= D (p ).

b. How many units will the consumer want to buy when the priceis P25 per unit?

c. What is the rate of change of p = P 25?


Applications


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Applications

Marginal Analysis

Let C (x ), R (x ), and P (x ) represent, respectively, the total cost,

revenue, and profit from the production and sale of x items.The marginal cost at x , given by C (x ) is the approximate cost of the (x + 1)st item:

C (x )

≈C (x + 1)

−C (x ) or C (x + 1)

≈C (x ) + C (x ).

The marginal revenue at x , given by R (x )is the approximaterevenue from the (x + 1)st item:

R (x )

≈R (x + 1)

−R (x ) or R (x + 1)

≈R (x ) + R (x ).

The marginal profit at x , given by P (x ) is the approximate profitfrom the (x + 1)st item:

P (x )

≈P (x + 1)

−P (x ) or P (x + 1)

≈P (x ) + P (x ).


Applications


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Applications

Marginal Analysis

Example 7. Given

C (x ) = 62x 2 + 27, 500 and R (x ) = x 3 − 12x 2 + 40x + 10,

find each of the following.

a. Total profit P (x ),

b. Total cost, revenue, and profit from the production and sale of 50 units of the product

c. The marginal cost, revenue, and profit when 50 units areproduced and sold.


Applications


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Applications

Solution:

a. The total profit is given by

Total profit = P (x ) = R (x ) − C (x )= x 3 − 12x 2 + 40x + 10 − (62x 2 + 27, 500)= x 3

−74x 2 + 40x

−27, 490.

b. The total cost, revenue, and profit from the production and saleof 50 units of the product are given by:

C (50) = 62

·502 + 27, 500 = $182, 500

R (50) = 503 − 12 · 502 + 40 · 50 + 10 = $97, 010P (50) = R (50) − C (50) = 97, 010 − 182, 500 = −$85, 490.

There is a loss of $85,490 when 50 units are produced and sold.


Applications


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Applications

c.

C (x ) = 124x

R (x ) = 3x 2 − 24x + 40P (x ) = 3x 2 − 148x + 40.

Hence,

C (50) = 124(50) = $6, 200

R (50) = 3 · (502) − 24(50) + 40 = $6, 340P (50) = 3 · (502) − 148(50) + 40 = $140

Therefore, once 50 units have been made, the approximate costof the 51st unit is $6,200 and from the sale of the 51st unit anapproximat revenue of $6,340 is expected. So, from theproduction and sale of the 51st item, an approximate profit of $140 will be gained.


Applications


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pp

Exercise 7a. Given

R (x ) = 50x

−0.5x 2 and C (x ) = 10x + 3,


a. Total profit P (x ),b. Total cost, revenue, and profit from the production and sale of

40 units of the product

c. The marginal cost, revenue, and profit when 40 units areproduced and sold.

Exercise 7b. Given

R (x ) = 5x and C (x ) = 0.001x 2 + 1.2x + 60,

find each of the following.a. Total profit P (x ),b. Total cost, revenue, and profit from the production and sale of

100 units of the productc. The marginal cost, revenue, and profit when 100 units are

produced and sold.Prof. Kenneth James T. Nuguid Calculus Applied to Business and Economics

Applications


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pp

Because of economies of scale and other factors, it is common forthe cost, revenue (price), and profit for, say, the 10th item to differ

from those for the 1000th item. For this reason, a business is ofteninterested in the average cost, revenue, and profit associated withthe production and sale of x items.

Average Cost, Revenue and Profit

Let C (x ), R (x ), and P (x ) be the cost, revenue and profitfunctions for producing and selling x items, then

C (x )

x = average cost of producing x items,

R (x )x

= average revenue for selling x items, and

P (x )

x = average profit for selling x items.


Applications


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pp

Application: Average Cost, Revenue and Profit

Example 8. Ken’s Greenhouse finds that the cost (in pesos) of growing x hundred sunflowers is modeled by

C (x ) = 20, 000 + 10, 000 4√

x .

If the revenue from the sale of x hundred sunflowers is given by

R (x ) = 12, 000 + 9, 000√

x ,


a. The average cost, the average revenue, and the average profitwhen x hundred sunflowers are grown and sold.

b. The rate at which average profit is changing when 300sunflowers are being grown and sold.


Applications


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pp

Solution:a. We let AC , AR , and AP represent average cost, average

revenue, and average profit, respectively. Then

AC (x ) =

C (x )

x =

20, 000 + 10, 000 4√

x

x ;

AR (x ) = R (x )

x =

12, 000 + 9, 000√

x

x ;

AP (x ) = P (x )

x

= R (x ) − C (x )

x

= −8, 000 + 9, 000√ x − 10, 000 4√ x

x

.


Applications


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b. To find the rate at which average profit is changing when 300

sunflowers are being grown, we calculate A

P (3) (remember thatx is in hundreds:

AP (3) = d

dx

−8, 000 + 9, 000x 1/2 − 10, 000x 1/4

x

x =3

= 4, 500x 1/2 − 2, 500x 1/4 + 8, 000 − 9, 000x 1/2 + 10, 0001/4

x 2

x

= 7, 500x 1/4 − 4, 500x 1/2 + 8, 000

x 2

x =3

= 7, 500 4√ 3 − 4, 500√ 3 + 8, 000

32

= P 1, 119.59


Applications


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Exercise 8. Sparkle Pottery has determined that the cost, in pesos,of producing x vases is given by

43, 000 + 21x 0.6.

If the revenue from the sale of x vases is given by R (x ) = 650x 0.9,find the rate at which the average profit per vase is changing when50 vases have been made and sold.


Chain Rule and Higher Derivatives


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Extended Power Rule

Suppose that g (x ) is a differentiable function of x . Then, for any

real number n,

d

dx [g (x )]n = n[g (x )]n−1 · d

dx g (x ).

Example 9. Find y

if y = (1 + x 2

)3

. Solution: Using the ExtendedPower Rule with g (x ) = 1 + x 2, we have

y = 3(1 + x 2)2 · d dx

(1 + x 2)

= 3(1 + x 2

)2

· 2x = 6x (1 + x 2)2.

Exercise 9a. Differentiate: f (x ) = (1 + x 3)1/2 Exercise 9b.Differentiate: f (x ) = (1

−x 2)3 + (5 + 4x )2




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Chain Rule

The derivative of the composition f ◦ g is given byd

dx [(f

◦g )(x )] =

d

dx [f (g (x ))] = f (g (x ))

·g (x ).

The Chain Rule often appears in another form. Suppose thaty = f (u ) and u = g (x ). Then

dy

dx =

dy

du ·du

dx .




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Example 10. Suppose y = 2 +√

u and u = x 3 + 1. Finddy

du

,

du

dx

, and dy

dx

. Solution:

dy

du =

1

2u −1/2 and

du

dx = 3x 2.

Then

dy

dx =

dy

du · du

dx

= 1

2√

u · 3x 2

= 3x 2

2√

x 3 + 1Substituting u = x 3 + 1.

Exercise 10. If y = u 2 + u and u = x 2 + x , find dy dx

.




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Application: Chain Rule

Example 11. A total-revenue function is given by

R (x ) = 1, 000

x 2 − 0.1x ,

where R (x ) is the total revenue, in thousands of dollars, from thesale of items. Find the rate at which total revenue is changing

when 20 items have been sold.

Solution: R (x ) = rate of change of revenue with respect to thenumber of items sold.

R (x ) = 1, 000

12

(x 2 − 0.1x )−1/2 · d dx

(x 2 − 0.1x )

= 500 · (2x − 0.1)√

x 2 − 0.1x .




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Hence,

R (20) = 500 · (2(20) − 0.1)

202 − 0.1(20)= 1, 000.

Example 11a. A company determines that its total cost, inthousands of pounds, for producing items is C (x ) =

√ 5x 2 + 60

and it plans to boost production t months from now according tothe function x (t ) = 20t + 40. How fast will costs be rising 4months from now?

Example 11b. If P10,000 is invested at interest rate i , compoundedquarterly, in 5 yr it will grow to an amount, A, given by

A(i ) = 10, 000

1 +

i

4

20.

Find the rate of change, dAdi

and interpret.




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Second Derivative

Given a differentiable function f (x ). Suppose that its derivative

function f

(x ) can also be differentiated. Then we can define thesecond derivative f ”(x ) as the function given by

f ”(x ) = d

dx f (x ).

Hence, it is clear that f ” is the derivative of f (x ).

Example12. Let f (x ) = x 5 − 3x 4 + x . Its second derivative isfound by differentiating f (x ).

f ”(x ) = d dx f (x )

= d

dx (5x 4 − 12x 3 + 1)

= 20x 3

−36x 2.




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Exercise 12a. If y = 1x , what is y ”.Exercise 12b. If y = (x 2 + 10x )20, find y and y ”.Exercise 12c. A company determines that monthly sales S , inthousands of dollars, after t months of marketing a product isgiven by

S (t ) = 2t 3 − 40t 2 + 220t + 160.

a. Find S (t ) and S ”(t ).

b. Find the value in (a.) when t = 1,

2,

4. Interpret the results.


Calculus Applied to Business Part04

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