Question

Part 2: Short answer questions Question 1 (4 points): A sausage firm has a production function of the form: q = 5LK+K+L where q is units per day, L is units of labor input and K is units of capital output. The marginal product of the two inputs are: MPL = 5K+1, MPK = 5L +1. Price per unit of labor: w= $15, price per unit of capital: v= $15. Both labor and capital are variable. a. Write down the condition for cost minimization. Explain in words. b. The firm wants to produce 2,040 units. What is the cost-minimizing combination of labor and capital for the firm to use? c. Using this optimal bundle of inputs, what will be the firm's total cost (C) to produce 2,000 units? d. Does this firm's production function exhibit constant returns to scale, increasing returns to scale or decreasing returns to scale?

Answer 1

Question 1 A sausage has a production for ction of the form: q: SLK + k +L where q= units pu day L= labor in put Ka capital out but Marginal product of labor, MPL- 2 2 2 = 5k41 Marginal Product of capital , MPx ² 29 = 52+1 Now Price of Pu hit of labor (w) = $15 : ... price pu mit of capital i = $15 Now Cost function : e = wzroka 152 + 15k ak . as condition for cost mini misation is i slope. if isoquent = Slope of isocost line ov MPL = wil MPK.. :!.-) 52+1 5K+1 5kti SLAI - =;) 5K+13 5271 a) 5K = 52.5) kol CS Scanned with . CamScanner

now, from from wants to produce 2040 mits. by now, je që 2040. Hace production function is: 2040=52kt kth Now from optimal input min Condition we get! KEL. subs tifuting K²2 in the production friction we git's 2040 - SLK + K+L -> 2040 = 5(2XL) + 2+2 -). 2040 523 +22 ej 522422 - 2040 = 0 Now by using qual quadratic formula weget! 2= - b ± √b24ac. .. za ...", and e= - 2040 Hue a=s, b=2 . Hence, ill - 27 V (2)²- 4*5*(-2040) 2.5. - 2 I r4+2060 40800 · : LE 10 1-21 2064 40804 20, -20.4 = T EMO - labor . is labor cannot be negative, this input to demand is 20 units Cs See Asd K =2, this k= 20 units, CamScanner

Huce cost minimizing combination of labor and capital for the firm to use is L= 20 mits and k = 20 uits. Production function es Now for 2000 mits bois. 2000SLK + k +2 input min condition Substituting optimal k = we get! 2000 = 5kLxu) + L+L 2) 2000 - 522 +22 2) 522+27:- 2000 20 Now by using quadratic formula weget! ..obt v beyae LETA 2a and C= - 2000 6=2 ne a=5, : kurce L = -b7 12)-- 485 x 6-2000) 2X5 .):23- -21 v 40004 cs Scanned with a 19, 80, - 20.20 CamScanner

as labor cannot be negative then optimal and as k=L L= 19.80 , this k = 19.80 Now, firms to tal cost (e) to produce 2000 mits will be found by substituting. L= 19.80 and ka 19.80 in the post function. les 152 +. 15k - (15 x 19.80)+(1S X 19.80) . = 4 594 d Now increasing both L. and k by a scalar constant a wc out? q(12sk) = 5(2) Ok) +6k) 7.al =5722k + k + X2 This production faction do not enhibits eithes, constant, increasing or decreasing ritmas to scale as it is not possible s to bring a out of the production fruiction, . . . . . Scanned With CamScanner