# Eco 507 Midterm

**Topics:**Supply and demand, Marginal cost, Consumer theory

**Pages:**6 (795 words)

**Published:**March 24, 2013

Midterm Test

1.(i.) ∝ =∆lnQ/∆lnP

∝ =P/Q* (∆Q/∆K) = Elasticity

The coefficients of double log model are the corresponding elasticities Price elasticity = 1.247

Income elasticity = 1.905

(ii.)Price elasticity = -1.2

Income elasticity = 2

Cross price elasticity = 1.5

Current volume = 10 mil

Average income increase by 2.5%

New qty after increase in income =

Ie=2

2=%∆Q%∆I

2=%∆Q/2.5

%∆Q=5%

New Qty = 11.445 mil

To increase the sales volume only by 9.2% you would have to reduce the price. %∆Q/%∆P=Pe

-5.25%∆P=-1.2

%∆P=4.375%

(iii). a. Maximize…Z = M + .5S + .5MS - S²

Subject to 30000S + 60000M = 1200000

Lagrangean…L=M+.5S+.MS-S2+λ1,200,000-30,000S-60,000M

∂L∂S=0.5+0.5M-2S-30,000λ

∂L∂M=1+0.5S-60,000λ

∂L∂λ=30,000S+60,000M

Equating λ, I get

1 + 0.5S/60000 = 0.5 + 0.5M – 2S

M = 4.5S

By substituting into budget constraint, I get

30000S + 60000 * 4.5S = 1200000

S = 4

M = 18

b. Cost function = 30000S + 60000M

Marginal cost of S = 30000

Marginal cost of M = 60000

Total marginal cost = 90000

c.

(iv.) a. Demand…Q = a – bP

E = (P/Q)*(∆Q/∆P)

E = -b (P/Q)

-0.4 = -b(4/2)

b = 0.2

a = Q + bP

= 2 + 0.2 * 4

a = 2.08

Demand Equation…Q = 2.08 – 0.2P

2.(i) Q = LK

∂Q∂L = K

∂2Q∂L2 = 0

The second order derivative did not give a negative value, so it ignores the condition of diminishing marginal productivity of labor.

b. Q (L, K) = LK

Q (mL, mK) = m²LK

The output increases more than proportionally, there are increasing returns to scale.

c.

Q = LK

TC = wL + rK

L = wL + rK + λ (Q-LK)

∂L∂L = w + λ (K) =0

∂L∂K = r + λ (L) =0

w /r = K/L =RTS

In this equation, the firm should use K and L as given that ratio to minimize cost of production. The Lagrangean Multiplier is marginal cost of any input to marginal benefit of any input should be same for any input. It explains if marginal cost –benefit ratio is greater for K than L, we have to substitute L for K to minimize cost.

d.

225 = LK

225 = 16L+144K

L = 16L+144K + λ (225-LK)

∂L∂L = 16 + λ (K) =0

∂L∂K = 144 + λ (L) =0

K/L =0.11

K = 0.11 L

L (0.11L) = 225

0.11 L^2 = 225

L^2= 2045.46

L = 45.23

45.23K = 225

K = 4.97

TC = 16*45.23+144*4.97

TC = $1439.36

e.

(ii) X dollars increase in the daily rate above $60, there are x units vacant. So 60+X= 80-X

2X=20

X=10

If they charge 60+10=$70, 10 rooms will be vacant and 70- rooms will be occupied. The profit for 80 rooms occupation at $60 per room,

TR= 80*60= $4800

TC= 4*80= $320

Profit = $4480

The profit for 70 rooms at the price of $70

TR= 70*70= $4900

TC= 4*70= $280

Profit= $4900 -$280= $4620

In this case the profit will also be maximized.

3. i)

a)

Maximize Y = 2Ty - .001Ty^2

S.t. 100Ty + 25Tz = 1300

Also

Maximize Z= 20 Tz - .01 Tz^2

S.t 100Ty + 25Tz = 1300

b) I used the Lagrangean to get:

L = 2Ty - .001Ty^2 + 20 Tz - .01 Tz^2

+θ (1300 - 100 Ty- 25Tz)

dL/dTy = 2 - 0.002Ty - θ(100) = 0

dL/dTz = 20 – 0.02Tz -θ(25) = 0

Also 100Ty + 25Tz = 1300

Divide the first two equation to get :

2 - 0.002Ty = θ(100)

20 – 0.02Tz =θ(25)

2- 0.002Ty = 100

20- 0.02Tz = 25

2-0.002Ty /20- 0.002Tz = 4

2- 0.002Ty = 80 – 0.008Tz

0.008 Tz – 0.002Ty = 78

100Ty + 25Tz = 1300

So T*y = 2.28 and Tz = 42.88

ii)

a) Q= 10 L – 0.1L ^2

Wage rate = 12

Now Q = 250 Then L required

Then L* = 50

And Labor price is 12 so total cost = 12x50 = 600 < 500. You should not accept the offer

b) Optimal amount of labor will be the one that equates MPL with wage ratio MPL = 10 – 0.2L = 2

8 = 0.2 L

L* = 40

And wage paid = 80

This is the optimal point and I should accept the offer as 80 < 500 Profit = 500 – 80 = 420

iii) To calculate the optimal price I used the markup formula that says that P – MC/ P = - 1/ed

Put the values to get

P- 10/P = -1/1.5

1.5 P – 15 = -P

2.5 P = 15

P* = 6...

Please join StudyMode to read the full document