Sorry, I replied too fast.
I think the answer is that you have to help Maxima out a bit here.
Perhaps something like this:
(%i31) solve([eq1,eq2],[X1,X2,lambda]);
(%o31) [[X1 = %r6,X2 = %r6*b*w1/(a*w2),lambda =
%r6^(-b-a+1)*a^(b-1)*w1^(1-b)*w2^b/b^b]]
(%i32) subst(%[1],eq3);
(%o32) Y-%r6^a*(%r6*b/a)^b*w1^b/w2^b = 0 <<< Maxima needs help
to solve this
(%i33) %-Y;
(%o33) -%r6^a*(%r6*b/a)^b*w1^b/w2^b = -Y
(%i34) radcan(%);
(%o34) -%r6^(b+a)*b^b*w1^b/(a^b*w2^b) = -Y <<< collect the %r6 terms
(%i35) solve(%,%r6);
Is b+a an integer?
y;
(%o35) [%r6 = (a^b/b^b)^(1/(b+a))*w2^(b/(b+a))*Y^(1/(b+a))/w1^(b/(b+a))]
(%i36) subst(%,%o31);
(%o36) [[X1 = (a^b/b^b)^(1/(b+a))*w2^(b/(b+a))*Y^(1/(b+a))/w1^(b/(b+a)),X2
= b*(a^b/b^b)^(1/(b+a))*w1^(1-b/(b+a))*w2^(b/(b+a)-1)*Y^(1/(b+a))/a,
lambda =
a^(b-1)*((a^b/b^b)^(1/(b+a)))^(-b-a+1)*w1^(-(-b-a+1)*b/(b+a)-b+1)*w2^((-b-a+1)*b/(b+a)+b)*Y^((-b-a+1)/(b+a))/b^b]]
(%i37) radcan(%); <<< perhaps a better form? perhaps not?
(%o37) [[X1 = (a^b)^(1/(b+a))*w2^(b/(b+a))*Y^(1/(b+a))/((b^b)^(1/(b+a))*w1^(b/(b+a))),
X2 = (a^b)^(1/(b+a))*b*w1^(a/(b+a))*Y^(1/(b+a))/(a*(b^b)^(1/(b+a))*w2^(a/(b+a))),
lambda =
a^(b-1)*(a^b)^(1/(b+a))*((a^b)^(1/(b+a)))^(-b-a)*((b^b)^(1/(b+a)))^(b+a)*w1^(a/(b+a))*w2^(b/(b+a))
/(b^b*(b^b)^(1/(b+a))*Y^((b+a-1)/(b+a)))]]
On Mon, Dec 13, 2010 at 21:08, Stavros Macrakis <macrakis at alum.mit.edu> wrote:
> solve([eq1,eq2,eq3],[X1,X2,lambda,Y]);
>
> Good luck in your adventure!
>
> ? ? ? ? ? ?-s
>
> On Sun, Nov 21, 2010 at 18:07, tomasz kopczewski <tomasze at aster.pl> wrote:
>> kill(all)$
>> assume(X1>0, X2>0,w1>0,w2>0,Y>0,lambda>0)$
>> L:(w1*X1+w2*X2)+lambda*(Y -(X1^a)*(X2^b));
>> eq1:diff(L,X1,1)=0;
>> eq2:diff(L,X2,1)=0;
>> eq3:diff(L,lambda,1)=0;
>> solve([eq1,eq2,eq3],[X1,X2,lambda]);
>