i'm not sure whether this y is given.cause if it is known, we could use it to transform the extremum with a constraint into an extremum without a constraint instead of using Lagrange
my answer is K=L=10√y~~

I didn't get what you said... Obviously, y could be regarded as given variable. But y is unknown. The question is to minimize the cost in the sense that y, w and r are given, L and K can be freely chosen.

First you should derive the minimizer L(y,w,r) and K(y,w,r). They are functions of y, w and r.

Then insert the f(L,K)=LK/100 and w=r=$1 into it and see the result.

That's all!

if y is given ,then Lagrange may be not necesary.

can you judge my answer is right or wrong?

OK....I give you my answer:

Define Lagrange function:

l = wL + rK - lambda * (f(L,K)-y)    PS. I will supress L and K in f

the minimizer L* and K* must satisfy the FOC:

dl/dL = w - lambda* df/dL =0   ==> w = lambda* df/dL

dl/dK = r - lambda* df/dK =0   ==> r = lambda* df/dK

dl/dlambda = f - y =0   ==> f = y

Now we have the general form of L* and K*, let f = LK/100. We have

df/dL = K/100  (1)

df/dK =L/100   (2)

substitution gives:

K* = 10*sqrt(wy/r)

L* = 10*sqrt(ry/w)

You get the right answer, but I think your lecturer prefers to see all the details.

thanks very much.