EXERCISE 3-7 Things to remember: 1. MARGINAL COST, …

EXERCISE 3-7

Things to remember:

1. MARGINAL COST, REVENUE, AND PROFIT

If x is the number of units of a product produced in some time interval, then: Total Cost = C(x) Marginal Cost = C'(x) Total Revenue = R(x) Marginal Revenue = R'(x) Total Profit = P(x) = R(x) - C(x) Marginal Profit = P'(x) = R'(x) - C'(x)

= (Marginal Revenue) - (Marginal Cost)

Marginal cost (or revenue or profit) is the instantaneous rate of change of cost (or revenue or profit) relative to production at a given production level.

2. MARGINAL COST AND EXACT COST

If C(x) is the cost of producing x items, then the marginal cost function approximates the exact cost of producing the (x + 1)st item:

Marginal Cost

C'(x)

Exact Cost C(x + 1) - C(x)

Similar interpretations can be made for total revenue and total profit functions.

3. BREAK-EVEN POINTS

The BREAK-EVEN POINTS are the points where total revenue equals total cost.

4. MARGINAL AVERAGE COST, REVENUE, AND PROFIT

If x is the number of units of a product produced in some time interval, then:

C(x) Average Cost = C (x) = x

Cost per unit

Marginal Average Cost = C '(x)

R(x) Average Revenue = R (x) = x

Revenue per unit

Marginal Average Revenue = R '(x)

P(x) Average Profit = P (x) = x

Profit per unit

Marginal Average Profit = P '(x)

EXERCISE 3-7 135

1. C(x) = 2000 + 50x - 0.5x2

(A) The exact cost of producing the 21st food processor is:

C(21) - C(20) = 2000 + 50(21) - (21)2

-

# %2000

+

50(20)

"

(20)2

& (

2

$ %

2 '(

= 2829.50 - 2800

= 29.50 or $29.50

(B) C'(x) = 50 - x

!

!

C'(20) = 50 - 20 = 30 or $30

3. C(x) = 60,000 + 300x

(A) C (x) = 60, 000 + 300x = 60, 000 + 300 = 60,000x-1 + 300

x

x

C (500) = 60, 000 + 300(500) = 210, 000 = 420 or $420

500

500

(B) C!'(x) = -60,000x-!2 =

"60, 000 x2

C '(!500) =

"60, 000 (500)2

= -!0.24

or -$0.24

Iinstedrepcrreetaastiinogn:a!tAtthae

production level of 500 rate of 24? per frame.

frames,

average

cost

(C) Tahpepr!aovxeirmaagteelcyos$t42p0er-

frame $0.24

if 501 frames = $419.76.

are

produced

is

5. P(x) = 30x - 0.3x2 - 250, 0 x 100

(A) The exact profit from the sale of the 26th skateboard is: P(26) - P(25) = 30(26) - 0.3(26)2 - 250 - [30(25) - 0.3(25)2 - 250] = 327.20 - 312.50 = $14.70

(B) Marginal profit: P'(x) = 30 - 0.6x P'(25) = $15

7. P(x) = 5x - x2 - 450 200

P'(x)

=

5

-

x 100

(A) P'!(450) = 5 - 450 = 0.5 or $0.50 100

Interpretation: At a production level of 450 cassettes, profit is increasing at the rate of 50? per cassette.

(B) P'(750) != 5 - 750 = -2.5 or -$2.50 100

Interpretation: At a production level of 750 cassettes, profit is decreasing at the rate of $2.50 per cassette.

!

136 CHAPTER 3 LIMITS AND THE DERIVATIVE

9. P(x) = 30x - 0.03x2 - 750

Average profit: P (x) = P(x) = 30 - 0.03x - 750 = 30 - 0.03x - 750x-1

x

x

(A) At x = 50, P (50) = 30 - (0.03)50 - 750 = 13.50 or $13.50. 50

(B)

P '(x)

=

! -0.03 +

750x-2

=

-0.03

+

!750 x2

P '(50) = -0.03 +

750 (50)2

= -0.03! + 0.3 = 0.27 or $0.27; at a

production level of 50 mowe!rs, the average profit per mower is INCREASING at the rate of $0.27 per mower.

(C) The average!profit per mower if 51 mowers are produced is approximately $13.50 + $0.27 = $13.77.

11. x = 4,000 - 40p (A) Solving the given equation for p, we get 40p = 4,000 - x and p = 100 - 1 x or p = 100 - 0.025x 40 Since p 0, the domain is: 0 x 4,00

(B) R(x) = xp = 100x - 0.025x2, 0 x 4,000

(C) R'(x) = 100 - 0.05x; R'(1,600) = 100 - 80 = 20 At a production level of 1,600 radios, revenue is INCREASING at

the rate of $20 per radio.

(D) R'(2,500) = 100 - 125 = -25 At a production level of 2,500 radios, revenue is DECREASING at

the rate of $25 per radio.

13. Price-demand equation: x = 6,000 - 30p Cost function: C(x) = 72,000 + 60x

(A) Solving the price-demand equation for p, we get p = 200 - 1 x; domain: 0 x 6,000 30

(B) Marginal cost: C'(x) = 60 (C) Revenue function: R(x) = 200x - 1 x2; domain: 0 x 6,000

30 (D) Marginal revenue: R'(x) = 200 - 1 x

15

(E) R'(1,500) = 100; at a production level of 1,500 saws, revenue is INCREASING at the rate of $100 per saw.

R'(4,500) = -100; at a production level of 4,500 saws, revenue is DECREASING at the rate of $100 per saw.

EXERCISE 3-7 137

Hundreds of Thousands

(F)

Thousands

(G) Profit function: P(x) = R(x) - C(x) = 200x - 1 x2 - [72,000 + 60x] 30 = 140x - 1 x2 - 72,000 30

(H) Marginal profit: P'(x) = 140 - 1 x 15

(I) P'(1,500) = 140 - 100 = 40; at a production level of 1,500 saws, profit is INCREASING at the rate of $40 per saw. P'(3000) = 140 - 200 = -60; at a production level of 3,000 saws, profit is DECREASING at the rate of $60 per saw.

15. (A) Assume p = mx + b. We are given 16 = m?200 + b

and 14 = m?300 + b Subtracting the second equation from the first, we get

-100m = 2 so m = - 1 = -0.02 50

Substituting this value into either equation yields b = 20. Therefore,

P = 20 - 0.02x; domain: 0 x 1,000 (B) Revenue function: R(x) = xp = 20x - 0.02x2, domain: 0 x 1,000. (C) C(x) = mx + b. From the finance department's estimates,

m = 4 and b = 1,400. Thus, C(x) = 4x + 1,400.

(D)

Hundreds of units

(E) Profit function: P(x) = R(x) - C(x) = 20x - 0.02x2 - [4x + 1,400] = 16x - 0.02x2 - 1,400

138 CHAPTER 3 LIMITS AND THE DERIVATIVE

Thousands of dollars

(F) Marginal profit: P'(x) = 16 - 0.04x P'(250) = 16 - 10 = 6; at a production level of 250 toasters, profit is INCREASING at the rate of $6 per toaster.

P'(475) = 16 - 19 = -3; at a production level of 475 toasters, profit is DECREASING at the rate of $3 per toaster.

17. Total cost: C(x) = 24x + 21,900 Total revenue: R(x) = 200x - 0.2x2, 0 x 1,000

(A) R'(x) = 200 - 0.4x The graph of R has a horizontal tangent line at the value(s) of x where R'(x) = 0, i.e., 200 - 0.4x = 0 or x = 500

(B) P(x) = R(x) - C(x) = 200x - 0.2x2 - (24x + 21,900) = 176x - 0.2x2 - 21,900

(C) P'(x) = 176 - 0.4x. Setting P'(x) = 0, we have 176x - 0.4x = 0

or x = 440

(D) The graphs of C, R and P are shown below.

Break-even points: R(x) = C(x) 200x - 0.2x2 = 24x + 21,900

0.2x2 - 176x + 21,900= 0

x = 176 ? (176)2 " (4)(0.2)(21, 900) (quadratic formula) 2(0.2)

= 176 ? 30, 976 " 17, 520 0.4

! = 176 ? 13, 456 = 176 ? 116 = 730, 150

0.4

0.4

Th!us, the break-even points are: (730, 39,420) and (150, 25,500). x-intercepts for P: -0.2x2 + 17.6x - 21,900 = 0

!

! or 0.2x2 - 176x + 21,900 = 0

which is the same as the equation above.

Thus, x = 150 and x = 730.

EXERCISE 3-7 139

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