Question #254286

Suppose that the XYZ Corp. (which is a profit-maximizing enterprise) produces “gadgets” according to the following production function: Q = 300K + 100KL + 2000L - L^2 , where Q is the number of gadgets per year, K is the amount of capital (machines) that are used, and L is the number of workers employed per year. Gadgets sell for $100 each.

a) If XYZ has 10 machines, and workers cost $100,000/year (including benefits and direct ancillary expenses), how many workers should XYZ hire?

b) If the workers cost $120,000, how many workers should XYZ hire?

c) If workers cost $100,000 but gadgets sell for $80 each, how many workers should XYZ hire?

d) If XYZ instead has 20 machines, how many workers should it hire when workers cost $100,000 and gadgets sell for $100?


1
Expert's answer
2021-10-21T10:50:40-0400

a.

Q=300K+100KL+2000LL2K=10Q=300(10)+100(10)L+20000LL2Q=3000+1000L+2000LL2MPL=dQdL=1000+20002L=30002LQ=300K+100KL+2000L-L^2\\K=10\\Q=300(10)+100(10)L+20000L-L^2\\Q=3000+1000L+2000L-L^2\\MP_L=\frac{dQ}{dL}=1000+2000-2L\\=3000-2L

At profit maximization, MPL=wpMP_L=\frac{w}{p}

30002L=10000010030002L=10002L=2000L=10003000-2L=\frac{100000}{100}\\3000-2L=1000\\2L=2000\\L=1000

1000 workers will be hired.


b.

Now w=120000

30002L=12000010030002L=1200L=9003000-2L=\frac{120000}{100}\\3000-2L=1200\\L=900

900 workers will be hired.


c.

Now w=100000 and p=80

30002L=10000080L=8753000-2L=\frac{100000}{80}\\L=875

875 workers will be hired.


d.

K=20

Q=3000(20)+100(20)L+2000LL2Q=60000+2000L+2000LL2MPL=dQdL=2000+20002L=40002L40002L=100000100L=1500Q=3000(20)+100(20)L+2000L-L^2\\Q=60000+2000L+2000L-L^2\\MP_L=\frac{dQ}{dL}=2000+2000-2L=4000-2L\\4000-2L=\frac{100000}{100}\\L=1500

1500workers will be hired



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