## Description

**Set 4**

**3.4: 1-17 odds3.6: 1-11 odds3.7: 1-21 odds4.1: 15-27 odds, 54-71odds4.2: 25-75 odds**

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## Explanation & Answer

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3.4

1

3.

𝐶(𝑥) = 2000 + 2𝑥 − 0.0001𝑥 2

(0 ≤ 𝑥 ≤ 6000)

a. 𝐶(1001) and 𝐶(2001)

𝐶(1001) = 2000 + 2(1001) − 0.0001(1001)2

𝐶(1001) = 2000 + 2002 − 100.2001

𝐶(1001) = 3901.7999

𝐶(2001) = 2000 + 2(2001) − 0.0001(2001)2

𝐶(2001) = 2000 + 4002 − 400.4001

𝐶(2001) = 5601.5999

b. 𝑀𝐶(1000) and 𝑀𝐶(2000)

𝑀𝐶 = 𝐶 ′ (𝑥)

𝑀𝐶 = 𝐶 ′ (𝑥) = (0) + 2(1) − 2(0.0001𝑥)

𝑀𝐶 = 2 − 0.0002𝑥

𝑀𝐶(1000) = 2 − 0.0002(1000)

𝑀𝐶(1000) = 2 − 0.2

𝑀𝐶(1000) = 1.8

𝑀𝐶(2000) = 2 − 0.0002(2000)

𝑀𝐶(2000) = 2 − 0.4

𝑀𝐶(2000) = 1.6

5.

𝐶(𝑥) = 100𝑥 + 200,000

a. average cost function, 𝐶

𝐶=

𝐶(𝑥)

𝑥

𝐶=

100𝑥 + 200,000

𝑥

𝐶 = 100 +

200,000

𝑥

b. marginal average cost function, 𝐶 ′

𝐶 ′ = (0) + 200,000(−1)(𝑥 −2 )

𝐶′ = −

200,000

𝑥2

c. As 𝑥 increases and becomes very large, the total cost also increases and becomes very large. However,

as the total cost and 𝑥 become very large, the average cost or cost per unit becomes smaller and smaller

and approaches zero.

7.

𝐶(𝑥) = 2000 + 2𝑥 − 0.0001𝑥 2

(0 ≤ 𝑥 ≤ 6000)

a. average cost function, 𝐶̅

𝐶̅ =

𝐶̅ =

𝐶(𝑥)

𝑥

2000 + 2𝑥 − 0.0001𝑥 2

𝑥

𝐶̅ =

2000

+ 2 − 0.0001𝑥

𝑥

̅

b. marginal average cost function, 𝐶′

̅̅̅

𝐶 ′ = 2000(−1)(𝑥 −2 ) + (0) − 0.0001(1)

̅ = − 2000 − 0.0001

𝐶′

𝑥2

9.

𝑅(𝑥) = 8000𝑥 − 100𝑥 2

a. marginal revenue, 𝑅 ′

𝑅 ′ = 8000(1) − 2(100𝑥)

𝑅 ′ = 8000 − 200𝑥

b. 𝑅 ′ (39), 𝑅 ′ (40), 𝑅 ′ (41)

𝑅 ′ (39) = 8000 − 200(39) = 200

𝑅 ′ (40) = 8000 − 200(40) = 0

𝑅 ′ (41) = 8000 − 200(41) = −200

c. Revenue is maximized when the marginal revenue, 𝑅′, is equal to 0. Therefore, the price should be

equal to $40 in order to maximize the revenue.

11.

𝑝 = −0.04𝑥 + 800

(0 ≤ 𝑥 ≤ 20,000)

𝐶(𝑥) = 200𝑥 + 300,000

a. profit function, 𝑃

𝑅(𝑥) = 𝑝𝑥

𝑅(𝑥) = (−0.04𝑥 + 800)𝑥

𝑅(𝑥) = −0.04𝑥 2 + 800𝑥

𝑃 = 𝑅(𝑥) − 𝐶(𝑥)

𝑃 = −0.04𝑥 2 + 800𝑥 − (200𝑥 + 300,000)

𝑃 = −0.04𝑥 2 + 600𝑥 − 300,000

b. marginal profit function, 𝑃′

𝑃′ = 2(−0.04𝑥) + 600(1) − (0)

𝑃′ = −0.08𝑥 + 600

c. 𝑃′ (5000) and 𝑃′ (8000)

𝑃′ (5000) = −0.08(5000) + 600 = 200

𝑃′ (8000) = −0.08(8000) + 600 = −40

d. Graph of 𝑃′

P

$ 1,950,000

518

7,500

14,483

x

From the graph, we see that the maximum profit that can be earned is equal to $ 1,950,000 when 7,500

units are produced. For the profit to be greater than zero, the number of units that should be produced

should be between 518 and 14,483 units.

13.

𝑝 = 600 − 0.05𝑥

(0 ≤ 𝑥 ≤ 12,000)

𝐶(𝑥) = 0.000002𝑥 3 − 0.03𝑥 2 + 400𝑥 + 80,000

a. revenue function, 𝑅, and profit function, 𝑃

𝑅 = 𝑝𝑥

𝑅 = (600 − 0.05𝑥)𝑥

𝑅 = 600𝑥 − 0.05𝑥 2

𝑃 = 𝑅 − 𝐶(𝑥)

𝑃 = 600𝑥 − 0.05𝑥 2 − (0.000002𝑥 3 − 0.03𝑥 2 + 400𝑥 + 80,000)

𝑃 = −0.000002𝑥 3 − 0.02𝑥 2 + 200𝑥 − 80,000

b. marginal cost function, 𝐶 ′ , marginal revenue function, 𝑅 ′ , and marginal profit function, 𝑃′

𝐶 ′ (𝑥) = 3(0.000002𝑥 2 ) − 2(0.03𝑥) + 400(1) + (0)

𝐶 ′ (𝑥) = 0.000006𝑥 2 − 0.06𝑥 + 400

𝑅 ′ = 600(1) − 2(0.05𝑥)

𝑅 ′ = 600 − 0.1𝑥

𝑃′ = 3(−0.000002𝑥 2 ) − 2(0.02𝑥) + 200(1) − (0)

𝑃′ = −0.000006𝑥 2 − 0.04𝑥 + 200

c. 𝐶 ′ (2000), 𝑅 ′ (2000), 𝑃′ (2000)

𝐶 ′ (2000) = 0.000006(2000)2 − 0.06(2000) + 400 = 304

𝑅 ′ (2000) = 600 − 0.1(2000) = 400

𝑃′ (2000) = −0.000006(2000)2 − 0.04(2000) + 200 = 96

From the calculations, we see that the marginal cost or additional cost gained when 2000 units are

produced is equal to $304, the marginal cost or additional revenue gained when 2000 units are sold is

equal to $400, and the marginal profit or additional profit gained when 2000 units are sold is equal to $96.

We see that the marginal profit is also equal to the difference of the marginal revenue and the marginal

cost, the same way that the profit is equal to the difference of the revenue and the cost.

d.

P

(6,000, $1,800,000)

(3,334, $290,370.37)

$ 80,000

419

5,872

x

12,000

In the graph, the red curve represents the cost function, the blue curve represents the revenue function,

and the green curve represents the profit function. Since the profit is the difference between the revenue

and cost, the profit curve is below both the cost and revenue curves. From the graph, we see that the

maximum revenue is equal to $1,800,000 when 6,000 units are produced, and the maximum profit is

equal to $290,370.37 when 3,334 units are produced. For the revenue to be greater than zero, the

company must sell between 0 and 12,000 units. For the profit to be greater than zero, the company must

sell between 419 and 5,872 units. The marginal cost, marginal revenue, and marginal profits represent

the slopes at a given point of their corresponding curve at a certain x value. For

example, 𝐶 ′ (2000), 𝑅 ′ (2000), 𝑎𝑛𝑑 𝑃′ (2000) are the slopes of the cost, revenue, and profit curves,

respectively, when 2000 units are produced.

15.

𝐶(𝑥) = 0.000002𝑥 3 − 0.03𝑥 2 + 400𝑥 + 80,000

a. average cost function, 𝐶̅

𝐶̅ =

𝐶(𝑥)

𝑥

𝐶̅ =

0.000002𝑥 3 − 0.03𝑥 2 + 400𝑥 + 80,000

80,000

= 0.000002𝑥 2 − 0.03𝑥 + 400 +

𝑥

𝑥

b. marginal average cost function, 𝐶̅ ′

𝐶̅ ′ = 2(0.000002𝑥) − 0.03(1) + (0) + 80,000(−1)(𝑥 −2 )

𝐶̅ ′ = 0.000004𝑥 − 0.03 −

80,000

𝑥2

c. 𝐶̅ ′ (5000) and 𝐶̅ ′ (10,000)

𝐶̅ ′ (5000) = 0.000004(5000) − 0.03 −

80,000

= −0.0132

(5000)2

𝐶̅ ′ (10000) = 0.000004(10000) − 0.03 −

80,000

= 0.0092

(10000)2

These two values represent the additional average cost when 5000 and 10000 units are produced,

respectively.

d.

𝐶̅

(6,000, $1,800,000)

(7827, $508,298)

x

17.

𝑝=

50

0.01𝑥 2 + 1

(0 ≤ 𝑥 ≤ 20)

a. revenue function, 𝑅

𝑅 = 𝑝𝑥

𝑅=(

𝑅=

50

)𝑥

0.01𝑥 2 + 1

50𝑥

0.01𝑥 2 + 1

b. marginal revenue function, 𝑅 ′

𝑅′ =

(0.01𝑥 2 + 1)50 − 50𝑥(0.02𝑥)

(0.01𝑥 2 + 1)2

𝑅′ =

0.5𝑥 2 + 50 − 𝑥 2

(0.01𝑥 2 + 1)2

𝑅′ =

−0.5𝑥 2 + 50

(0.01𝑥 2 + 1)2

c. 𝑅 ′ (2)

𝑅 ′ (2) =

−0.5(2)2 + 50

= 44.38

(0.01(2)2 + 1)2

From the calculation, we can interpret 𝑅 ′ (2) as the additional revenue when 2 units are sold, which is

equal to $44.38.

3.6

1.

𝑥 + 2𝑦 = 5

a.

2𝑦 = 5 − 𝑥

𝑦=

5−𝑥

2

𝑦 = 2.5 − 0.5𝑥

𝑑𝑦

= (0) − 0.5

𝑑𝑥

𝑑𝑦

= −0.5

𝑑𝑥

b.

𝑥 + 2𝑦 = 5

𝑑

𝑑

(𝑥 + 2𝑦) =

(5)

𝑑𝑥

𝑑𝑥

1+2

2

𝑑𝑦

=0

𝑑𝑥

𝑑𝑦

= −1

𝑑𝑥

𝑑𝑦

= −0.5

𝑑𝑥

3.

𝑥𝑦 = 1

a.

𝑦=

1

𝑥

𝑑𝑦

= −1(𝑥 −2 )

𝑑𝑥

𝑑𝑦

1

=− 2

𝑑𝑥

𝑥

b.

𝑥𝑦 = 1

𝑑

𝑑

(𝑥𝑦) =

(1)

𝑑𝑥

𝑑𝑥

𝑥

𝑑𝑦

+ (1)𝑦 = 0

𝑑𝑥

𝑥

𝑑𝑦

= −𝑦

𝑑𝑥

𝑑𝑦

𝑦

=−

𝑑𝑥

𝑥

𝑊𝑒 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑦 =

𝑑𝑦

1 1

=− ( )

𝑑𝑥

𝑥 𝑥

1

𝑥

𝑑𝑦

1

=− 2

𝑑𝑥

𝑥

5.

𝑥 3 − 𝑥 2 − 𝑥𝑦 = 4

a.

−𝑥𝑦 = 4 − 𝑥 3 + 𝑥 2

𝑦=

4 − 𝑥3 + 𝑥2

−𝑥

4

𝑦 = − + 𝑥2 − 𝑥

𝑥

𝑑𝑦

= −4(−1)(𝑥 −2 ) + 2(𝑥) − (1)

𝑑𝑥

𝑑𝑦

4

= 2 + 2𝑥 − 1

𝑑𝑥 𝑥

b.

𝑑 3

𝑑

(𝑥 − 𝑥 2 − 𝑥𝑦) =

(4)

𝑑𝑥

𝑑𝑥

3𝑥 2 − 2𝑥 − (𝑥

−𝑥

𝑑𝑦

+ 𝑦) = 0

𝑑𝑥

𝑑𝑦

= −3𝑥 2 + 2𝑥 + 𝑦

𝑑𝑥

𝑑𝑦 −3𝑥 2 + 2𝑥 + 𝑦

=

𝑑𝑥

−𝑥

4

𝑊𝑒 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑦 = − + 𝑥 2 − 𝑥

𝑥

4

2

2

𝑑𝑦 −3𝑥 + 2𝑥 − 𝑥 + 𝑥 − 𝑥

=

𝑑𝑥

−𝑥

𝑑𝑦

4

= 3𝑥 − 2 − 2 − 𝑥 + 1

𝑑𝑥

𝑥

𝑑𝑦

4

= − 2 + 2𝑥 − 1

𝑑𝑥

𝑥

7.

𝑥

− 𝑥2 = 1

𝑦

a.

𝑥

= 1 + 𝑥2

𝑦

𝑦=

𝑥

1 + 𝑥2

𝑑𝑦 (1 + 𝑥 2 )(1) − (𝑥)(2𝑥)

=

(1 + 𝑥 2 )2

𝑑𝑥

𝑑𝑦 1 + 𝑥 2 − 2𝑥 2

=

(1 + 𝑥 2 )2

𝑑𝑥

𝑑𝑦

1 − 𝑥2

=

𝑑𝑥 (1 + 𝑥 2 )2

b.

𝑑 𝑥

𝑑

(1)

( − 𝑥2) =

𝑑𝑥 𝑦

𝑑𝑥

𝑦(1) − 𝑥 (

(

𝑑𝑦

)

𝑑𝑥

𝑦2

𝑦(1) − 𝑥 (

𝑦2

𝑦(1) − 𝑥 (

−𝑥 (

) − 2𝑥 = 0

𝑑𝑦

)

𝑑𝑥

= 2𝑥

𝑑𝑦

) = 2𝑥𝑦 2

𝑑𝑥

𝑑𝑦

) = 2𝑥𝑦 2 − 𝑦

𝑑𝑥

𝑑𝑦 2𝑥𝑦 2 − 𝑦

=

𝑑𝑥

−𝑥

𝑑𝑦

𝑦

= −2𝑦 2 +

𝑑𝑥

𝑥

𝑊𝑒 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑦 =

𝑥

1 + 𝑥2

𝑥

(

)

2

𝑑𝑦

𝑥

1 + 𝑥2

= −2 (

)

+

𝑑𝑥

1 + 𝑥2

𝑥

𝑑𝑦

−2𝑥 2

1

=

+

2

2

𝑑𝑥 (1 + 𝑥 )

1 + 𝑥2

𝑑𝑦 −2𝑥 2 + (1 + 𝑥 2 )

=

(1 + 𝑥 2 )2

𝑑𝑥

𝑑𝑦

1 − 𝑥2

=

𝑑𝑥 (1 + 𝑥 2 )2

9.

𝑥 2 + 𝑦 2 = 16

𝑑 2

𝑑

(𝑥 + 𝑦 2 ) =

(16)

𝑑𝑥

𝑑𝑥

2𝑥 + 2𝑦 (

2𝑦 (

𝑑𝑦

)=0

𝑑𝑥

𝑑𝑦

) = −2𝑥

𝑑𝑥

𝑑𝑦

2𝑥

=−

𝑑𝑥

2𝑦

𝑑𝑦

𝑥

=−

𝑑𝑥

𝑦

11.

𝑥 2 − 2𝑦 2 = 16

𝑑 2

𝑑

(𝑥 − 2𝑦 2 ) =

(16)

𝑑𝑥

𝑑𝑥

2𝑥 − 2(2𝑦)

−4𝑦

𝑑𝑦

=0

𝑑𝑥

𝑑𝑦

= −2𝑥

𝑑𝑥

𝑑𝑦

2𝑥

=−

𝑑𝑥

−4𝑦

𝑑𝑦

𝑥

=

𝑑𝑥 2𝑦

3.7

1.

𝑓(𝑥) = 2𝑥 2

𝑑𝑦

= 4𝑥

𝑑𝑥

𝑑𝑦 = 4𝑥𝑑𝑥

.

3.

𝑓(𝑥) = 𝑥 3 − 𝑥

𝑑𝑦

= 3𝑥 2 − 1

𝑑𝑥

𝑑𝑦 = (3𝑥 2 − 1)𝑑𝑥

5.

1

𝑓(𝑥) = √𝑥 + 1 = (𝑥 + 1)2

1

𝑑𝑦 1

1

= (𝑥 + 1)−2 (1) =

𝑑𝑥 2

2√𝑥 + 1

𝑑𝑦 = (

1

2√𝑥 + 1

) 𝑑𝑥

7.

3

1

𝑓(𝑥) = 2𝑥 2 + 𝑥 2

1

1

𝑑𝑦 3

1

= (2𝑥 2 ) + (𝑥 −2 )

𝑑𝑥 2

2

1

𝑑𝑦

1

= 3𝑥 2 + 1

𝑑𝑥

2𝑥 2

1

1

𝑑𝑦 3𝑥 2 (2𝑥 2 ) + 1

=

1

𝑑𝑥

2𝑥 2

𝑑𝑦 6𝑥 + 1

=

1

𝑑𝑥

2𝑥 2

6𝑥 + 1

𝑑𝑦 = (

1 ) 𝑑𝑥

2𝑥 2

9.

𝑓(𝑥) = 𝑥 +

2

𝑥

𝑑𝑦

= (1) + 2(−1)(𝑥 −2 )

𝑑𝑥

𝑑𝑦

2

=1− 2

𝑑𝑥

𝑥

𝑑𝑦 𝑥 2 − 2

=

𝑑𝑥

𝑥2

𝑑𝑦 = (

𝑥2 − 2

) 𝑑𝑥

𝑥2

11.

𝑓(𝑥) =

𝑥−1

𝑥2 + 1

𝑑𝑦 (𝑥 2 + 1)(1) − (𝑥 − 1)(2𝑥)

=

(𝑥 2 + 1)2

𝑑𝑥

𝑑𝑦 𝑥 2 + 1 − 2𝑥 2 + 1

=

(𝑥 2 + 1)2

𝑑𝑥

𝑑𝑦

2 − 𝑥2

= 2

𝑑𝑥 (𝑥 + 1)2

𝑑𝑦 = (

2 − 𝑥2

) 𝑑𝑥

(𝑥 2 + 1)2

13.

1

𝑓(𝑥) = √3𝑥 2 − 𝑥 = (3𝑥 2 − 𝑥)2

1

𝑑𝑦 1

= (3𝑥 2 − 𝑥)−2 (6𝑥 − 1)

𝑑𝑥 2

𝑑𝑦

=

𝑑𝑥

6𝑥 − 1

2(3𝑥 2 −

𝑑𝑦 = (

1

𝑥)2

=

6𝑥 − 1

2√3𝑥 2 − 𝑥

6𝑥 − 1

2√3𝑥 2 − 𝑥

) 𝑑𝑥

15.

𝑦 = 𝑓(𝑥) = 𝑥 2 − 1

a.

𝑑𝑦

= 2𝑥 − (0) = 2𝑥

𝑑𝑥

𝑑𝑦 = 2𝑥𝑑𝑥

b. 𝑥: 1 → 1.02

∆𝑥 = 1.02 − 1 = 0.02

∆𝑦 =

𝑑𝑦

∆𝑥

𝑑𝑥

∆𝑦 = 2(1)(0.02) = 0.04

c.

𝑓(1) = (1)2 − 1 = 0

𝑓(1.02) = (1.02)2 − 1 = 0.0404

𝑓(1.02) − 𝑓(1) = 0.0404 − 0 = 0.0404

The two values are almost equal with their difference being equal to 0....