Maximum Value

Thnx prophet sir.

Method 2: Fix x and substitute for z and by taking derivative w.r.t y you will see that f decreases with y

Now fix x and substitute for y and taking derivative w.r.t z we note that f increases up until x+2z<2.

So we maximize f by y→0 and x+z=1 and x+2z→2 which is achieved when we further have x→0 and z→1 as above

You have an upper bound of 4, a maximum cannot be found

Method 1:

Note that 0< x,y,z <1.

Let f(x,y,z) = (1-x)(2-y)(3-z)

Now its easy to verify that since y<1, we have

f(x,y,z) < f(0,y+x,z)

Further since z<1, we have

f(0,y,z) < f(0,0,z+y)

Note that in the above transformations, the sum of the variables remains unchanged

From the above two inequalities and since 0<x,y,z<1 we can maximize f x,y \rightarrow 0^+ and z\rightarrow 1^- and hence f→4

ya

ya..ur argument is OK...

in that case ans should be as given by anirudh

@akari r u not getting ans by lagranges multiplier?

Subjective dear.

This is a Qsn from FIITJEE....So I think the soln is well within JEE range....Ur method's outta JEE syllabus.