Question

Suppose a firm’s production function is given by Q = L^1/2*K^1/2. The Marginal Product of Labor and the Marginal Product of Capital are given by:

MP_L = ½ L^-1/2K^1/2and MP_K = ½ L^1/2K^-1/2

a) Suppose the price of labor is w = 18, and the price of capital is r = 2. Derive the firm’s total cost function.

b) What is the firm’s marginal cost?

c) For this problem, you will sketch the graph of the firm’s isoquant for Q = 12 units of output, and on the same graph sketch the firm’s isocost line associated with the total cost of producing Q = 12 units of output. To get this total cost, you must use the Total Cost function from part a). Please scale your graph up to 72 units of Labor on the horizontal axis, and 72 units of Capital on the vertical axis (do not go above 72 units on either axis). For the isocost line, clearly identify the vertical and horizontal intercepts. For the isoquant, clearly identify 4 combinations of Labor and Capital that will produce Q = 12 (including the bundle that minimizes the firm’s cost of production). Make sure your graph is neatly and accurately drawn and carefully labeled.

Answer 1

A firm's production function is given by ! a) o = 1/2K/2 labor . Marginal product of MPL = 20 = 2 2 ² 4/2 L2L. Marginal Roduct of capital MPK = da I L2K-²2 Now Price of labor w= 18, price of capital na 2. Now cost function ez. WLtok. -=) es 182 +2K Now Optimal in put mix condition is MP =) { [-%k% Ź Luz k=%% = 12 2 € = 9 -> K=9L substituting k=qL in the production Now bum ction we gut? Scanned with CamScanner

Talla 32 Q = 1/2 k % -) Q = _Y2 (94)'h a 20 a) D=L and as; K= 9L= q* 9 = 3Q Now, e substituting 2 = 0 and K= 3Q in the Cost function we get, essit e 2.184 +2K. =) C= (18 * @) + 2 * (32) . =) e 60+ 6Q :-) ė : 129 - Huce to tal cost fonction e=120 6) Marginal cost bustion me. = de la = 12 e Now, for out but Q = 7 12 mits; . L a = 12 = 4 and ka 30 213 X 12). 3 . z 36. Hnce cost minimising buble is .. (L, k) = (4, 36) Scanned with CamScanner

. Now when Q=12 Iso quart equation is 12=2%2K / 2 KM :-) When 12 = k½ = L = 4 .2=4 =) 140/2 =) 12 = K² 1 . -) K12=6 =) k= 36 -) K whi L= q :=) 9/ =.9 :=) 12 = k½ akk -) k=4 =) K= 16 ..=) 1 2 2 = K² Whis L=16 :) . =) kk = 3 = 9 . =) k wm L=36 .=) 12 = K² 36% K2=2. =) K-4 cs Hace 4 combinations of Land K ore 14,136), (9,16); (16, 9) ad. (36,4) Camscanner

Now why Q = 12 Total cost ez 12 Q = (12x12) = 144 tice equation of iso cost lines : e= WLtok =) 144 = 181+2K Now horizontal in Ancept of the isocost line [ substitute K=0] . er : 144 = 182+ (2x0) =) L=8 Now untical in Ancept of the iso cost line [ substitute 2=07 . .244 = (18 x0) +2% -) K = 72 - Now by Joining horizontal and uutical in Incept we get the iso cost line El AB and the iso quart q= 12 will be tagurt to AB at cost minimising bundle (L) k) = (4,36) CS Scanned with .CamScanner

727 A 36% 14,36) 2012 n 0 : 4 8 9 72 L CS Scanned with CamScanner

Answer 2

It is solved below:

W = 18 0=2 Short vun total cost =WLtolk = 18L+QK - (2) sowing foort from equation (!-> 2 Putting this value in equation as STO = 18(Q2 ) +ak. ASTO = 0 = - 180 +2=0 = - 18Q" = -242 => K²=9Q2 27 K = 30 Now, long oun total cost => 1826 ) +2.36 - 6Q + 6 Q = 12.9 Total cost : C = 120

B. Marginal cost = dc = 12 d = 12

C = 120 212.12 = 144 Total cost when 9 = 12 = Q = 12, K= 36 = 36 La H (4, 36) is one combination 36 that prodreces 12 unite State * If K = 40 L = 36; Q=12. SO, (36, 4) is an other combination

If L=16, K = 9, 6 = 12 +2 = 4.3=12 (16,9) is the 3rd combination Similarly 9, 16) is the 4th combination Cost = WL + DIK N=18, de=2 For the 1st bundles, G= 18.4+2.36=72772 For the and bundle, c = 18.36 +2.4 = +2.4 = 648+8 = 656 648+8 .

For the 3rd bundle , c = 18.16+2.9=288+18=306 For the 4th bandie, e=18.9 +2.16=162+32 = 194 so, the 1st bundle 7 (L,K) = (4, 36) minimizes frorm's cost of poroduction (144).

367 (4,36) (9, 1G) \(16,9) (36,4) 4 ģ 16 3 0 L intercept of the isocost line must be L=144, K=0 and K=141, L=0. only then 0212 would be possible