Question

5. For this problem you will analyze the elasticity of substitution and the isoquant graphs for the following production function F(K.L) = 2K1/2 1/2 (a) Graph the isoquant for F(K,L) that represents an output of 10. Be sure to show your work and label the axes clearly. Also, two points should be clearly labelled. (3 points) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 (b) What is the marginal rate of technical substitution for F? (2 points) Screenshot

Answer 1

Q = 2K^1/2 L^1/2

Put Q = 10

10 = 2K^1/2 L^1/2

(10 / 2)= K^1/2 L^1/2

5 = K^1/2 L^1/2

(5 /  L^1/2 ) = K^1/2

(5 /  L^1/2 )² = (K^1/2)²

(25 / L) = K (Isoquant equation for output level of 10)

(25 /L ) =K
L	K
1	25.00
2	12.50
3	8.33
4	6.25
5	5.00
6	4.17
7	3.57
8	3.13
9	2.78
10	2.50
11	2.27
12	2.08
13	1.92
14	1.79
15	1.67
16	1.56
17	1.47
18	1.39
19	1.32
20	1.25

20.00 19.00 18.00 Capital 900 N 5. K=5 on 1 =10. K-2.5 0.00 0 1 2 3 4 5 6 7 8 11 12 13 14 15 16 17 18 19 20 9 10 Labor

(b) Marginal Rate of Technical Substitution (MRTS) = (MPL / MPK)

$MP L = \Delta F(K,L)/ \Delta L$

MPL = 2 * (1/2) K^1/2 L^-1/2

MPL = K^1/2 L^-1/2

$MP K = \Delta F(K,L)/ \Delta K$

MPK= 2 * (1/2) K^-1/2 L^1/2

MPK= K^-1/2 L^1/2

MRTS = MPL /MPK

MRTS = ( K^1/2 L^-1/2 / K^-1/2 L^1/2 )

MRTS = K / L

(c)

Elasticity of substitution (o): o= -O(K/L) MRTS OMRTS (K/L)

we have, MRTS = K/L

AMRTS/ACK/L) = 1

$\Delta (K/L)/\Delta MRTS = 1$

Elasticity of substitution = 1 * (MRTS / MRTS) (Note: MRTS = K/L)

Elasticity of substitution = 1