# Math Help - 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? 2. 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. 3. ## 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... 4. Ooops. My mistake. Didn't convert. You are correct. 900.8 5. ## 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? 6. 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

7. 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.