Answer
The maximum value of $f$ subject to the line $x+y=1$ is $\frac{{{a^a}{b^b}}}{{{{\left( {a + b} \right)}^{a + b}}}}$.
Work Step by Step
We are given $f\left( {x,y} \right) = {x^a}{y^b}$ for $x \ge 0$, $y \ge 0$, where $a,b > 0$ are constants.
Our task is to find the maximum value of $f$ subject to the constraint $g\left( {x,y} \right) = x + y - 1 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {a{x^{a - 1}}{y^b},b{x^a}{y^{b - 1}}} \right) = \lambda \left( {1,1} \right)$
So, the Lagrange equations are
(1) ${\ \ \ \ }$ $a{x^{a - 1}}{y^b} = \lambda $, ${\ \ \ }$ $b{x^a}{y^{b - 1}} = \lambda $
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
We have $x \ge 0$, $y \ge 0$. Since $\left( {0,0} \right)$ does not satisfy the constraint, we have the following cases:
Case 1. $x=0$, $y \ne 0$
Substituting $x=0$ in the constraint gives $y=1$. So, the critical point is $\left( {0,1} \right)$.
Case 2. $x \ne 0$, $y=0$
Substituting $y=0$ in the constraint gives $x=1$. So, the critical point is $\left( {1,0} \right)$.
Case 3. $x \ne 0$, $y \ne 0$
In this case, equation (1) implies that $\lambda \ne 0$. So,
$\lambda = a{x^{a - 1}}{y^b} = b{x^a}{y^{b - 1}}$
$\frac{{{y^b}}}{{{y^{b - 1}}}} = \frac{b}{a}\frac{{{x^a}}}{{{x^{a - 1}}}}$
$y = \frac{b}{a}x$
Step 3. Solve for $x$ and $y$ using the constraint
In Step 2, we obtain $y = \frac{b}{a}x$. Substituting it in the constraint gives
$x + \frac{b}{a}x - 1 = 0$
$x\left( {\frac{{a + b}}{a}} \right) = 1$
So, $x = \frac{a}{{a + b}}$.
Using $y = \frac{b}{a}x$, we obtain $y = \frac{b}{{a + b}}$. So, the critical point is $\left( {\frac{a}{{a + b}},\frac{b}{{a + b}}} \right)$.
Step 4. Calculate the critical values
We evaluate the extreme values at the critical points and list them in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }point}}}&{f\left( {x,y} \right)}\\
{\left( {0,1} \right)}&0\\
{\left( {1,0} \right)}&0\\
{\left( {\frac{a}{{a + b}},\frac{b}{{a + b}}} \right)}&{\frac{{{a^a}{b^b}}}{{{{\left( {a + b} \right)}^{a + b}}}}}
\end{array}$
From the results in this table, and since $a,b > 0$, we conclude that the minimum of $f$ is $0$ and the maximum value of $f$ subject to the line $x+y=1$ is $\frac{{{a^a}{b^b}}}{{{{\left( {a + b} \right)}^{a + b}}}}$.
