Question

1. Dirt Diggers (DD) is a firm that excavates roadside ditches to lay drainpipe. Its output follows the production function: Q = 10L – .1L², where Q denotes the length of the ditch in meters. The firm hires labor at a wage of $12 per hour. DD has received an offer to excavate 250 meters for a price of $500. Should it accept the offer? Suppose DD is offered as much or as little excavation work as it desires at a price of $2.00 per meter, what is the optimal quantity that it should choose to excavate?

MP_L = dQ/dL = 10 – .2L

Answer 1

Q = 10L -.1L2

250 = 10L - 0.1L^2

L = 50

Total Cost = 50*$12 = $600

Offer = $500

Profit = 500-600 = -$100

Profit is negative. So, should not take the offer.

Offer = $2/meter

Total Cost = 12L

Total Revenue = 2Q = 20L - .2L^2

Profit = 20L - .2L^2 - 12L > 0

Profit is positive so it should take.

For optimum profit level,

dP/dL = 8 - .4L = 0

L = 20

Profit = 8L - .2L2 = 80

Q = 10L -.1L2 = 160

So optimum quantity is 160 m.