Assuming [tex]f(x,y)=x^ay^b[/tex]. We have Lagrangian
[tex]L(x,y,\lambda)=x^ay^b+\lambda(x+y-1)[/tex]
with partial derivatives (set to 0)[/tex]
[tex]L_x=ax^{a-1}y^b+\lambda=0\implies ax^{a-1}y^b=-\lambda[/tex]
[tex]L_y=bx^ay^{b-1}+\lambda=0\implies bx^ay^{b-1}=-\lambda[/tex]
[tex]L_\lambda=x+y-1=0[/tex]
[tex]\implies ax^{a-1}y^b=bx^ay^{b-1}\implies -bx+ay=0[/tex]
[tex]x+y-1=0\implies x+y=1[/tex]
Solving this system of linear equations yields [tex]x=\dfrac a{a+b}[/tex] and [tex]y=\dfrac b{a+b}[/tex] as the sole critical point, which in turn gives a maximum value of [tex]f\left(\dfrac a{a+b},\dfrac b{a+b}\right)=\dfrac{a^ab^b}{(a+b)^{a+b}}[/tex].
