# find x and z which maximizes..

• Jan 26th 2007, 02:16 PM
chogo
find x and z which maximizes..
does anyone know how to do this

find x and z that maximizes ln(4-x) + ln(4+z) + ln(z)

subject to

2 lnx + ln(4-z) + ln(4-z) = 2ln(32/7)

anyhelp is much appreciated
• Jan 26th 2007, 02:51 PM
CaptainBlack
Quote:

Originally Posted by chogo
does anyone know how to do this

find x and z that maximizes ln(4-x) + ln(4+z) + ln(z)

subject to

2 lnx + ln(4-z) + ln(4-z) = 2ln(32/7)

anyhelp is much appreciated

the values of x and z that maximise ln(4-x) + ln(4+z) + ln(z), also
maximise:

(16-x^2)z ...(1)

as x>0, the constraint may be rewritten:

x^2(4-z)^=(32/7)^2,

or:

x=32/(7(4-z)) ...(2)

So substitute (2) into (1) and find the z that maximises this now
unconstrained objective, find the corresponding x from 2, and substitute
back into ln(4-x) + ln(4+z) + ln(z) to find its maximum.

RonL
• Jan 26th 2007, 02:55 PM
chogo
thank you very much
• Jan 26th 2007, 04:36 PM
chogo
im sorry to be a pain but i seem to be very thick and dont understand how you obtained equation (1)

(16-x^2)z ...(1)

if you have the time could you please explain this to me
• Jan 26th 2007, 10:25 PM
CaptainBlack
Quote:

Originally Posted by chogo
im sorry to be a pain but i seem to be very thick and dont understand how you obtained equation (1)

(16-x^2)z ...(1)

if you have the time could you please explain this to me

You are looking for the maximum of g(x,z)=ln(4-x) + ln(4+z) + ln(z), but:

g(x,z)=ln( (4-x)(4+x)x )

by the law of logarithms.

Then let:

f(x,z)=exp(g(x,z))=(4-x)(4+x)x,

and because the exponential function is increasing on the real numbers, any
maxima of g(x,z) is also a maxima of f(x,z) and vice-versa.

RonL