Approximation by Increments

[Becky4paws]

At a factory, the daily output is Q(L)=60,000L^1/3 units, where L denotes the size of the labor force measured in worker hours. Crrrently 1,000 worker hours are used each day. Estimate the effect on output that will be produced if the labor force is cut to 940 worker hours.

Q(L) = 60,000L^1/3
Q'(L) = 20,000L^-2/3

L=1,000 and the change of L = -60

Q(1000) = 60,000(1,000)^1/3=60,000(1/10)=60,000/10=6,000
Q'(1,000) = 20,000(1000)^-2/3=20,000(1/100)??? = 20,000/100 = 200

Then applying (change of Q)(Q'(1000)) + Q(1000)

 

[soroban]

Hello, Becky4paws!

At a factory, the daily output is \(\displaystyle Q(L)\:=\:60,000L^{\frac{1}{3}}\),
where \(\displaystyle L\) denotes the size of the labor force measured in worker hours.
Currently 1,000 worker hours are used each day.
Estimate the effect on output if the labor force is cut to 940 worker hours.

We are expected to use differential approximation.

The differential of the function is: \(\displaystyle \,dQ\:=\:20,000L^{-\frac{1}{3}}\,dL\;=\;\frac{20,000\,dL}{L^{\frac{2}{3}}}\)

We are given: \(\displaystyle \,L\,=\,1000\,\) and \(\displaystyle \,dL\,=\,-60\)

Therefore: \(\displaystyle \,dQ\:=\:\frac{(20,000)(-60)}{1000^{\frac{2}{3}}} \:=\:-12,000\)

That is, about 12,000 fewer units will be produced daily.


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We can check the accuracy of our approximation.

The original output is: \(\displaystyle \,Q(1000)\:=\:60,000\left(1000^{\frac{1}{3}}\right) \:=\:600,000\)

The new output is: \(\displaystyle \,Q(940)\:=\:60,000\left(940^{\frac{1}{3}}\right)\:\approx\:587,752\)

The actual change of output \(\displaystyle (\Delta Q)\) is: \(\displaystyle \,12,249\) fewer units.

So our approximation is quite accurate.