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scaling problems in "optim"

p.s.  With proper scaling, 'optim' gives the correct answer in this case: 

(mle2 <- optim(rep(0, 2), dmvnorm, method='CG',
               control=list(fnscale=-10, trace=9),
              mean=1:2, hessian=TRUE, log=TRUE))
$par
[1] 0.9999917 1.9999833

      However, as noted below, with parscale = c(0.3, 3), only a 
difference of a factor of 9 in the scaling between the two parameters, 
the first parameter = 0.735 rather than 1;  the second is identified 
correctly. 

      Spencer
Spencer Graves wrote: