# Thread: Log linear maximisation problem

1. ## Log linear maximisation problem

We are given to solve for t to maximise W,

W=(1-(1/(t+1)))^a*(1/(t+1))^b*e^-tE

where t>0 or t=0

Which can be log linearised to obtain

lnW=aLn(1-(1/(t+1)))+bLn(1/(t+1))-tE

Obviously to obtain the maximimum we need to differentiate the above with respect to t and find the zero value(s) for the differential, (and then locate which is the maximum) but I am stuck on how to proceed with the differentiation.

Any help would be most appreciated.

2. ## Re: Log linear maximisation problem

You need to differentiate:

$\displaystyle a\ln\left(1-\frac{1}{t+1}\right)+b\ln\left(\frac{1}{t+1}\right )-tE$

with respect to t. There are three terms. The third one is easy: $\displaystyle \frac{d}{dt}(-tE)=-E$.

The first:

$\displaystyle \frac{d}{dt}a\ln\left(1-\frac{1}{t+1}\right)$$\displaystyle = a\frac{1}{\left(1-\frac{1}{t+1}\right)}\frac{d}{dt}\left(\frac{-1}{t+1}\right) = a\frac{1}{\left(1-\frac{1}{t+1}\right)}\frac{1}{(t+1)^2} The second: \displaystyle \frac{d}{dt}b\ln\left(\frac{1}{t+1}\right)$$\displaystyle = b\frac{1}{\left(\frac{1}{t+1}\right)}\frac{d}{dt} \left(\frac{1}{t+1}\right)$$\displaystyle = -b\frac{1}{\left(\frac{1}{t+1}\right)}\frac{1}{(t+1 )^2}$

And I'll leave it to you to simplify the result.

You might be able to differentiate the original function, too, but this seems to give a simpler answer.

- Hollywood