marginal analysis

• Aug 9th 2006, 02:08 PM
becky
marginal analysis
At a factory, the daily output is Q(K) = 600K^1/2 units, where K is the capital investment measured in units of $1,000. The current capital investment is$900,000. Estimate the effect that an additional capital investment of $800 will have on the daily output. I'm not sure I am setting this up correctly. Q'(K) = 300K^-1/2 K=900,000/1,000 Change in K=0.8 Is the problem set up by finding Q'(900) and Q'(900.8) and then subtracting the two totals? • Aug 9th 2006, 02:31 PM Jameson Close but not completely right. You can use the slope of the output graph to estimate how much each additional unit(s) will cost close to 900. The more units you add/subtract the worse the estimate gets. This is where you use Q'(K). Let's think of linearization. $L(x)=f'(a)(x-a)+f(a)$ So our "a" value is 900, and our "x" value, our estimate, is 900.8. Take it from here. • Aug 9th 2006, 03:00 PM becky Need explaination please You put the x value at 908. I thought it would be 900.8. I rationalized it as 900,000/1,000 = 900, then if you add 800 it would be 900,800/1,000 = 900.8. Please help me see your way... • Aug 9th 2006, 03:03 PM Jameson Ooops. My mistake. Didn't convert. You are correct. 900.8 :eek: • Aug 9th 2006, 03:20 PM becky Then - here I go... L(x) =2/3(900.8-900)+270,000 L(x) = 2/3(.8) +270,000 L(x) = 1.6/3 +270,000 L(x) =270,000.533 I still don't feel confident with this answer. Any chance it's right? • Aug 9th 2006, 07:52 PM CaptainBlack Quote: Originally Posted by becky At a factory, the daily output is Q(K) = 600K^1/2 units, where K is the capital investment measured in units of$1,000. The current capital investment is $900,000. Estimate the effect that an additional capital investment of$800 will have on the daily output.

I'm not sure I am setting this up correctly.

Q'(K) = 300K^-1/2
K=900,000/1,000
Change in K=0.8

Is the problem set up by finding Q'(900) and Q'(900.8) and then subtracting the two totals?

$
\Delta Q\approx \frac{dQ}{dK} \Delta K= 300K^{-1/2} \Delta K
$

So here $K=900$, $\Delta K=0.8$, hence:

$
\Delta Q\approx 300\times 900^{-1/2}\times 0.8=8 \mbox{ units}
$

RonL
• Aug 10th 2006, 07:25 AM
Jameson
Sorry I misread your question a bit. My linearization formula gave you the change plus the original value to give the new estimated value. If you want just the change, use this part of the linearization formula.

$\Delta(x) \approx f'(a)(x-a)$

which is what CaptainBlack told you to do.