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Homework answers / question archive / The monthly cost function, in dollars, for a coffee maker factory is C(x)C(x) where xx is the number of coffee makers produced
The monthly cost function, in dollars, for a coffee maker factory is C(x)C(x) where xx is the number of coffee makers produced. The financial model used by management predicts that C(1000)=5000C(1000)=5000 and C′(1000)=10C′(1000)=10.
1. How much would you expect monthly cost to increase if production were increased from 10001000 to 10061006 coffeemakers?
2. Find the average cost per coffee maker at a production level of 10001000 coffeemakers per month.
3. Find the marginal average cost at a production level of 10001000 coffee makers.
Given Data
(a)
Production is increased from 1000 to1006
Therefore,
C(1006)=C(1000)+C′(1000)⋅(1006−1000)(assuming Euler method)C(1006)=5000+10⋅6C(1006)=5060C(1006)=C(1000)+C′(1000)⋅(1006−1000)(assuming Euler method)C(1006)=5000+10⋅6C(1006)=5060
ΔC=C(1006)−C(1000)=5060−5000=60dollarsΔC=C(1006)−C(1000)=5060−5000=60dollars
In other words, since the marginal cost is the increase in the cost of a product as a result of an extra unit of production,
then the increase in the cost of a product as a result of an extra 6 units of production can estimated by 6 times the marginal cost.
Monthly cost increase will be 60 dollars.
(b)
The expression for the average cost per coffee maker is,
AC=C(x)xAC=C(x)x
Substitute the known values,
AC=10x−5000x=10(1000)−50001000=10−5=5AC=10x−5000x=10(1000)−50001000=10−5=5
Thus, the average cost per coffee maker is 55.
(c)
The marginal average cost is,
MAC=ddx(AC)MAC=ddx(AC)
Substitute the known values,
MAC=ddx(10x−5000x)=0+5000(1x2)=(5000x2)MAC=ddx(10x−5000x)=0+5000(1x2)=(5000x2)
Substitute the known values,
MAC=(5000(1000)2)=0.005MAC=(5000(1000)2)=0.005
Thus, the marginal average cost is 0.0050.005.
