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CHAPTER 4

Question 1

Q=300S+200U-0.2S^2-0.3U^2

Ps=10

Pu=5

MPs=300-0.4S

MPu=200-0.6U

MPs/Ps=MPu/Pu

(300-0.4S)/10=(200-0.6U)/5

300-0.4S=400-1.2U

1.2U=0.4S+100

U=0.33S+83.33

If S= 400, then U=0.33(400)+83.33=216.66 If we want to maximize profits and use 400 hours of skilled labor, we would need 216.66 hours of unskilled labor. We would not follow Elwyn’s recommendation. 10S+5U=5000 U=1000-2S

U=0.33S+83.33

1000-2S=0.33S+83.33

2.33S=916.67

S=393.42

U=1000-2(393.42)=213.16

MR(MPU)=MEU 10(200-0.6U)=5

200-0.6U=0.5

U=(200-0.5)/(0.6)

U=332.5

Question 6

AP= total milk/total grain = (5917+7250+8379+9371)/(1200+1800+2400+3000) = 3.68 pounds of milk per pound of grain

MP between 1200 and 1800 = (7250-5917)/(1800-1200) = 2.22 MP between 1800 and 2400 = (8379-7250)/(2400-1800) = 1.88 MP between 2400 and 3000 = (9371-8379)/(3000-2400) = 1.65

Yes it does. Marginal product decreases as units of grain increase. CHAPTER 5 Question 3

TC= TVC+TFC TC=100Q+5000

Profit = TR – TC

10000=300Q – 100Q+5000

5000 = 200Q

Q=25

10000=350Q-100Q+5000 5000=250Q

Q=20

10000=350Q-85Q+5000 5000=265Q

Q=18.86

Around 19 units.

Question 4

When Q is 50, C=16.68+0.125(50)+0.00439(50)^2

C=33.905

When Q is 51,

C=16.68+0.125(51)+0.00439(51)^2

C=34.473

Difference of an increase of 0.568

The manager can know that the marginal cost is positive and can calculate the marginal cost. This can help him maximize profits with a bit more calculations and data. MC=0.125+0.00878Q

This can help managers maximize profits if they can make MC=MR. Question 5

Output Total Cost TFC TVC AFC AVC

0 50 50 0 - -

1 75 50 25 50 25

2 100 50 50 25 25

3 120 50 70 16.66 23.33

4 135 50 85 12.5 21.25

5 150 50 100 10 20

6 190 50 140 8.33 23.33

7 260 50 210 7.14 30

Question 6

TC=Q*AC TC=3Q+4Q^2

No it doesn’t because when Q=0, TC = 0 also. This means that all costs come from variable costs. MC=3+8Q

TR=3Q

MR=3

MC=MR

3+8Q=3

Q=0

The company is making a loss because their profit is maximized when Q=0. The more they sell and make, the more they lose. TC=3Q+4Q^2 MC=3+8Q

Question 9

MC=50-20Q+3Q^2 Will be at a minimum when the derivative of MC = 0 -20+6Q=0

Q=20/6=3.33

AVC=TVC/Q AVC=50-10Q+Q^2

Will be at a minimum when the derivative of AVC = 0

-10+2Q=0

Q=5

AVC=50-10(5)+5^2 =25

MC=50-20(5)+3(5^2)

=25

Question 10

TVC=$14 TFC=$300,000

For 10000 units, will cost

TC=TVC+TFC

=14*10000+300000

=440000

TR=25*10000

=250000

Profit=250000-440000

=-190000

Lose 190000$ and so, they should not introduce the new product. Profit = 200000-440000 =--240000

Lose 240000$ and so, they should not introduce the new product. Profit = 150000-440000 =-290000

Lose 290000$ and so, they should not introduce the new product. The minimum price they should charge is 44$ per unit if they can sell 10000 units. They would make 0$ profit in this case. Question 11 Degree=(23000+11000-30000)/30000=0.133

When production facilities are used to produce one product, sometimes another product is also created in the process. If this newly created product were to be used for the creation of other products, these other products would be cheaper to make. Question 12

AVC=$24 TVC=24/Q

24$=40%

Profit + FC = 60% of table

(24/4)*10=60$ per table

Last year, sold 10000 tables. if price increase by 10%, we will need to sell 10000/1.1=9090.91

Around 9091 tables.

((241.08)/410)/60=1.08 9090.91/1.08=8417.51

Around 8418 tables

CHAPTER 6

Question 1

AC=(250000/Q)+150+3Q MAC=(-25000/Q^2)+3=0...

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