Hi roadapples-ga,
I use the signs "<=" and ">=" to mean "less/greater than or equal to",
respectively.
Define f(a,b) = a^2 + (b-1)^2, and g(b) = e^(2b) + (b-1)^2. Then,
under the constraint e^b - a <= 0 (or, equivalently, a >= e^b), we
have
f(a,b) >= g(b)
So let us first minimize g(b) as a function of b. Differentiate:
g'(b) = 2e^(2b) + 2(b-1) = 2(e^(2b) + b - 1)
So g'(b)=0 if and only if e^(2b)+b-1 = 0, this happens for b=0, and
ONLY for b=0, since g''(b) = 4e^(2b)+2 is a strictly positive
function, so g' is increasing and can only hit zero once. Furthermore
since g''(0) is positive this means that g has a local minimum at 0,
but because it's the only critical point it's also a global minimum.
So we have shown:
g(b) >= g(0) = 2 for all b
and
f(a,b) >= g(b) >= 2 for all a,b satisfying the constraint.
But setting b=0, a=e^b=1 gives f(a,b)=2. This proves that
a=1, b=0 is the global minimum for f(a,b) under the constraint.
This is the answer.
Hope this helps, please ask for clarification if you have any doubts.
Regards,
dannidin |