if x+y=1 find min value of xlogx+ylogy?

Question

the ans is -log2 but let me know the steps of solution

Answer 1

Note by symmetry we expect it to be when x = y, so x = y = 1/2 and x log x + y log y = 1/2 (log 1/2) + 1/2 (log 1/2) = log (1/2) = -log 2. But that doesn't prove it. ;-)

Rewrite as x log x + (1-x) log (1-x) and differentiate:
(I'm assuming natural logs here, but if you want base b logs, it's just a constant factor of log b out the front, so it makes no difference):
x (1/x) + log x - (1-x) / (1-x) - log (1-x) = 0
=> log x = log (1-x)
=> x = 1-x
=> x = 1/2
so we get x = y = 1/2 and the value of -log 2 as expected.