Question

(4) use the method of lagrange multipliers to determine the maximum value of f(x, y) = x a y b (the a and b are two fixed positive constants) subject to the constraint x + y = 1. (assume x, y > 0.)

Answer 1

Assuming $f(x,y)=x^ay^b$. We have Lagrangian

$L(x,y,\lambda)=x^ay^b+\lambda(x+y-1)$

with partial derivatives (set to 0)[/tex]

$L_x=ax^{a-1}y^b+\lambda=0\implies ax^{a-1}y^b=-\lambda$
$L_y=bx^ay^{b-1}+\lambda=0\implies bx^ay^{b-1}=-\lambda$
$L_\lambda=x+y-1=0$

$\implies ax^{a-1}y^b=bx^ay^{b-1}\implies -bx+ay=0$
$x+y-1=0\implies x+y=1$

Solving this system of linear equations yields $x=\dfrac a{a+b}$ and $y=\dfrac b{a+b}$ as the sole critical point, which in turn gives a maximum value of $f\left(\dfrac a{a+b},\dfrac b{a+b}\right)=\dfrac{a^ab^b}{(a+b)^{a+b}}$.