Question

Suppose the wage rate offered by a firm is $10 per hour and the rental rate of capital is $25 per hour. (a) What does an isocost line measure? Write the equation for the isocost line for the firm. (b) Illustrate (with labour on the horizontal axis and capital on the vertical axis) the isocost line for C=$800. Label the horizontal and vertical intercepts on your diagram. Calculate and label the slope of the isocost line. (c) Suppose the price of capital falls to $20 per hour. Show what happens to C=$800 isocost line. Include any changes in intercepts and the slope. How would you interpret this change?

Answer 1

(a) Isocost line shows the different combination of labor and capital which cost same to the firm.

Let w is wage rate

=> w = $10

and

r is rental rate of capital  

=> r = $25

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Isocost line equation: wL + rK = C

where L is labor;. K is capital and C is cost

=>10L + 25K = C

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(b) C= 800 (given)

Isocost line equation: 10L + 25K = C
=> 10L + 25K = 800

L is measured on horizontal axis and K is measured on Vertical axis.

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Horizontal axis intercept:

10L + 25K = 800

Put K =0 and solve for L

=> 10L + 25(0) = 800

=> L = (800 /10)

=> L = 80

and

Vertical axis intercept

10L + 25K = 800

Put L =0 and solve for K

=> 10(0) + 25K= 800

=> K = (800 / 25)

=> K = 32

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Slope of isocost line = -(w/r)

=> Slope of isocost line = -(10 / 25)

=> Slope of isocost line = -0.4

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(c) now price of capital fall from $25 to $20.

New r = 20

Isocost line equation: wL + rK =C

Put w = 10; r=20 and C=800

=> 10L + 20K = 800

--------------------------

Horizontal axis intercept:

10L + 20K = 800

Put K =0 and solve for L

=> 10L + 20(0) = 800

=> L = (800 /10)

=> L = 80

There is no change in horizontal intercept of isocost line.

and

Vertical axis intercept

10L + 20K = 800

Put L =0 and solve for K

=> 10(0) + 20K= 800

=> K = (800 / 20)

=> K = 40

The intercept of vertical axis intercept changes from 32 to 40 due to fall in price of capital.

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As a result of the fall in price of capital, the isocost line rotated outside on the vertical axis.

Slope of new isocost line = -(w/r)

=Slope of new isocost line = -(10 / 20)

=> Slope of new isocost line = -0.5

Slope of isocost line change from -0.4 to -0.5