ECON 211 SPRING 2023 ASSIGNMENT #4 PRODUCTION AND COSTS

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ECON 211 SPRING 2023 ASSIGNMENT #4 PRODUCTION AND COSTS

Q1: (10 points) Let Q(L) = 12L2 – (L3)/4 with costs = 1000 + 50L.

a) (5 points) Complete the table on the next page. You can do this in excel or by hand.

TABLE: Column 1:  Labour input from 1-24

Column 2: Output resulting from L: Q(L)      =            12L2 – (L3 )/4

Column 3:  Average Product of Labour APL            =            Q/L

Column 4:  Marginal Product of Labour: MP        =            ∆Q/∆L

Column 5:  Cost: COST    =             1000 + 50L

Column 6:  Average Cost: AC         =            COST/Q

Column 7:  Average Cost: AVC      =            (COST- 1000)/Q

Column 8:  Marginal Cost: MC       =            ∆C/∆Q

b) (1 point) At what value of L is output maximized? (eg, the firm will never hire more than this L.)

For L=32, output is 4096.  If we raise labour, say to 33, then Q=4083.  Hence max output is

4096.  This means the firm will never produce past L=32 since it costs more but will reduce output.  It cannot be profit maximizing.

c) (1 point) Confirm that APL is maximized where MPL = APL. (You can plot APL and MPL versus L. Or you can use the table directly. Find the maximum APL. For labour below this we should see MPL > APL. For labour above we should see MPL < APL.)

Use:  scatter plot with L in first column and APL, MPL is second and third column.

Confirmed:  Maximum APL is at L = 24.  If L is less than this, MPL is higher than APL which raises APL .  For L larger than 24, MPL is smaller and so reduces APL .

d) (1 point) d) Confirm that AC is minimized where MC = AC : if you plot this, then just show in your plot.

Minimum AC occurs at Q=3920 (L=28).  If Q is less than this, MC is lower than AC which raises AC.  For Q larger than 3920, MC is larger and so raises AC.

e) (1 point) Confirm that: AVC is a minimum where APL is a maximum

Minimum AVC is 0.347 and occurs at Q=3456 (L=24).  APL is maximized at the same level.  The idea is that AVC= wL/Q = w/(Q/L) = w/APL    hence as APL  rises, AVC falls (and vice versa).

f) (1 point) Confirm that: MC is a minimum where MPL is a maximum

Minimum MC is 2.60 at Q=2018  with L = 16.  But this is where MPL is at a maximum of 192

Q2: (10 points) Consider a firm that employs capital and labour and uses the production technology Q = min[10K, 10L] (think of taxi cabs: you need exactly one car and one driver per      taxi but each taxi can generate 10 rides per shift). Factor prices are r for capital and w for labour.

a) (3 points) Draw the isoquant for Q = 100.

Recall:  an isoquant is combinations of K and L that produce Q=100.

ISSUE:  inputs here are perfect complements so the Isoquant is a ‘right-angle’ .  The corner is at L=10 and K=10.  That is,  we need to set 10K=10L or K=L.  for Q=100  we have K=100/10=10

b) (1 point) Does this production function exhibit increasing, constant, or decreasing returns to scale? Show/explain.

DEFN:  CRS implies a doubling of inputs will exactly double inputs.

So if we compare L=10 and K=10  with Q = 100 to L=20 and K=20  with Q=200, we see that we have CRS

c) (1 point) What is the optimal relationship between K and L? What is the input demand for Labour (L as a function of Output) and the input demand for Capital?

The optimal relationship is K = L:  we use inputs one for one.  Therefore input demands are K = Q/10  and L=Q/10

d) (2 point) Derive the long-run cost function: (eg C = wL + rK where L and K are optimal choices).

STEP 1:  we have to find C = wL + rK for and level of output Q

STEP 2:  for any Q we must have K = Q/10 and L = Q/10 C = wQ/10 + rQ/10

STEP 3:  so C = Q(w+r)/10

e) (1 point) What is the long-run marginal and average cost function with r = 10 and w = 5?

MC= ∆C/∆Q = (w+r)/10 = (10+5)/10 = $1.5

AC = COST/Q = (w+r)/10 = (10+5)/10 = $1.5

Notice that since AC is constant (not rising or falling) it must be equal to MC.

f) (1 point) What is the cost of Q=100? Q= 200?

C(Q=100) = Q (15/10) =  150

Eg:  to get Q=100, we need K=10 and L=10.  So wL + rK = 5(10)+10(10) = 150.

Alternately C= Q * AC = 100*1.5 = 150

C(Q=200) = 300

g) (1 point) Does this cost function exhibit increasing, constant, or decreasing economies of scale?

DEFN:  CES implies a doubling of output will exactly double costs.

Clearly doubling output does double costs.

Q3: (10 points) Consider the following production function: Q = K1/3 L1/3 with rental costs r and labour costs w.

a) (1 point) Draw the isoquant for Q = 100.

Want combinations of K and L that yield Q=100.  We approach this the same way that we did the Cobb- Douglas preferences.

STEP 2:  isolate L1/3   :                                                 L1/3 = 100/ K1/3

STEP 3:  raise both sides by power of 3            (L1/3 )3 = (100/ K1/3 )3

STEP 4: simplify:                                                         L = 1003/ K

STEP 5:  create table of K and L (and check that Q = 100)

b) (1 point) Does this production function exhibit increasing, constant, or decreasing returns to scale? Show/explain.

We have Decreasing Returns to Scale:

Step 1                 Set K = 100 and L = 10,000 so Q = 100 from table

Step 2                 Now double K and L:  Set K = 200 and L = 20,000  hence Q = 158 .

Step 3:               since Q increases but by less than double, we have DRS

c) (1 point) Let K be variable. What is the optimal relationship between K and L: (hint: MRTS = L/K)

Optimality (tangency between the isoquant and the isocost line) implies Y = rK/w

STEP 1 Optimality requires MRTS = r/w  where MRTS = MPK / MPL

STEP 2 MPK   =  ⅓ K1/3 – 1  L1/3  and MPL   = ⅓ K1/3 – 1  L1/3

STEP 3 so MRTS =  [⅓ K1/3 – 1  L1/3 ]/[ ⅓ K1/3 – 1  L1/3  ] = L/K

STEP 4: now set to = r/w L/K = r/w  or L = rK/w

d) (2 points) What is the input demand for Labour (as a function of Output) and the input demand for Capital?

Need to substitute the optimality condition back into the production function to get K as a function of Q and L as a function of Q

STEP 1 sub Y = rK/w Q  = K1/3  L1/3 =  K1/3  (rK/w) 1/3

STEP 2 simplify Q  =  K2/3  (r/w) 1/3

STEP 3 isolate K2/3 K2/3  = Q (w/r) 1/3

STEP 4 raise both sides to power (3/2) K* = Q3/2 (w/r) 1/2

This is the input demand for K:  if Q rises we need more K; if r rises we use less K; if w rises, we

use more K

STEP 5 sub into Y = rK/w L = (r/w) Q3/2  (w/r) 1/2

STEP 6 simplify L* = Q3/2 (r/w) 1/2

This is the input demand for L:  if Q rises we need more L; if r rises we use more L ; if w rises, we use less L

e) (1 point) Derive the long-run cost function. (eg C = wL + rK where L and K are optimal choices).

All we do is substitute L* and K* into C= wL* + rK*

STEP 1 substitute C = w[Q3/2  (r/w) 1/2 ] + r[Q3/2  (w/r) 1/2 ]

STEP 2 simplify C = Q3/2   r ½ w ½ + Q3/2   r ½ w ½

STEP 2 simplify more C = 2 Q3/2   r ½ w ½

Notice:  costs rise if Q rises, if r rises, or if w rises.

f) (1 point) Plot the cost function given w = $10 and r = $10.

From above:  C= 2 Q3/2  (10) ½ (20) ½ = 93.91

g) (1 point) Calculate the long-run costs for Q = 100. Q = 200.

See above table:  C(Q=100) = 20,000   and C(Q=200) = 56,568

h) (2 points) Does this cost function exhibit increasing, constant, or decreasing economies of scale? Why does this happen?

As we can see, a doubling of output more than doubles costs.  The reason is that we have DRS.   To double output we need to more than double inputs.  Eg to go from Q=100 to Q=200 we need to raise K from 1000 to 2828 and L from 1000 to 2828.

TOTAL: 30 points

2023-06-26