On my computer the solution of
var('a x y p_x p_y D Rev R l')
assume(a,'real')
assume(x,'real')
assume(y,'real')
assume(p_x,'real')
assume(p_y,'real')
assume(D,'real')
assume(Rev,'real')
assume(R,'real')
assume(l,'real')
assume(a<1)
assume(a>0)
assume(p_x>0)
assume(p_y>0)
assume(R>0)
U =(1/(a+1))*x^(a+1)+y
show(LatexExpr(r'\text{La fonction d}^\prime\text{utilité est }U(x,y) =
'),U)
D= x*p_x + y*p_y
show(LatexExpr(r'\text{La Dépense } D = '),D)
Rev= R
show(LatexExpr(r'\text{Le Revenu } Rev = '),R)
L=U+l*(Rev-D)
show(LatexExpr(r'\text{Le lagrangien est } \mathcal{L}(x, y, λ) = '),L)
FOC = [diff(L,x),diff(L,y),diff(L,l)]
show(LatexExpr(r'\text{Les condition du premier ordre sont }
\left\{\begin{array}{c}\mathcal{L}_x= 0\\\mathcal{L}_y= 0\\\mathcal{L}_λ=
0\end{array}\right. '))
show(LatexExpr(r'\text{soit }'))
show(LatexExpr(r'\mathcal{L}_x= 0 \Longleftrightarrow '),FOC[0]==0)
show(LatexExpr(r'\mathcal{L}_y= 0 \Longleftrightarrow '),FOC[1]==0)
show(LatexExpr(r'\mathcal{L}_λ= 0 \Longleftrightarrow '),FOC[2]==0)
sol=solve([FOC[0]==0,FOC[1]==0,FOC[2]==0],x,y,l)
show(sol)
Is nearly correct, but an extra complex exponential term multiplies $x$ and
then modifies $y$. Even as an element I do not understand its form :
$e^{(2iπz_{5797}a)}$
Could some one explain why ?
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