Answer
We show that the maximum value of $f\left( {x,y} \right) = {x^2}{y^3}$ on the unit circle ${x^2} + {y^2} = 1$ is $\frac{6}{{25}}\sqrt {\frac{3}{5}} $.
Work Step by Step
We are given $f\left( {x,y} \right) = {x^2}{y^3}$. Since the unit circle is ${x^2} + {y^2} = 1$, the constraint is $g\left( {x,y} \right) = {x^2} + {y^2} - 1 = 0$.
Our task is to find the maximum value of $f$ subject to the constraint $g\left( {x,y} \right) = {x^2} + {y^2} - 1 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {2x{y^3},3{x^2}{y^2}} \right) = \lambda \left( {2x,2y} \right)$
So, the Lagrange equations are
(1) ${\ \ \ \ }$ $2x{y^3} = 2\lambda x$, ${\ \ \ }$ $3{x^2}{y^2} = 2\lambda y$
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
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 $g\left( {x,y} \right) = {x^2} + {y^2} - 1 = 0$ gives $y = \pm 1$. So, the critical points are $\left( {0,1} \right)$, $\left( {0, - 1} \right)$.
Case 2. $x \ne 0$, $y=0$
Substituting $y=0$ in the constraint $g\left( {x,y} \right) = {x^2} + {y^2} - 1 = 0$ gives $x = \pm 1$. So, the critical points are $\left( {1,0} \right)$, $\left( { - 1,0} \right)$.
Case 3. $x \ne 0$, $y \ne 0$
In this case $\lambda \ne 0$. From equation (1) we obtain
$\lambda = {y^3} = \frac{3}{2}{x^2}y$
${y^2} = \frac{3}{2}{x^2}$
$y = \pm \sqrt {\frac{3}{2}} x$
Step 3. Solve for $x$ and $y$ using the constraint
In Step 2, we obtain $y = \pm \sqrt {\frac{3}{2}} x$. Substituting it in the constraint gives
${x^2} + {y^2} - 1 = 0$
${x^2} + \frac{3}{2}{x^2} = 1$
So, $x = \pm \sqrt {\frac{2}{5}} $.
Using $y = \pm \sqrt {\frac{3}{2}} x$, we obtain the critical points: $\left( {\sqrt {\frac{2}{5}} ,\sqrt {\frac{3}{5}} } \right)$, $\left( {\sqrt {\frac{2}{5}} , - \sqrt {\frac{3}{5}} } \right)$, $\left( { - \sqrt {\frac{2}{5}} ,\sqrt {\frac{3}{5}} } \right)$, $\left( { - \sqrt {\frac{2}{5}} , - \sqrt {\frac{3}{5}} } \right)$.
Step 4. Calculate the critical values
We evaluate the extreme values at the critical points and list them in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }point}}}&{f\left( {x,y} \right)}\\
{\left( {0,1} \right)}&0\\
{\left( {0, - 1} \right)}&0\\
{\left( {1,0} \right)}&0\\
{\left( { - 1,0} \right)}&0\\
{\left( {\sqrt {\frac{2}{5}} ,\sqrt {\frac{3}{5}} } \right)}&{\frac{6}{{25}}\sqrt {\frac{3}{5}} }\\
{\left( {\sqrt {\frac{2}{5}} , - \sqrt {\frac{3}{5}} } \right)}&{ - \frac{6}{{25}}\sqrt {\frac{3}{5}} }\\
{\left( { - \sqrt {\frac{2}{5}} ,\sqrt {\frac{3}{5}} } \right)}&{\frac{6}{{25}}\sqrt {\frac{3}{5}} }\\
{\left( { - \sqrt {\frac{2}{5}} , - \sqrt {\frac{3}{5}} } \right)}&{ - \frac{6}{{25}}\sqrt {\frac{3}{5}} }
\end{array}$
From the results in this table we conclude that the minimum of $f$ is $ - \frac{6}{{25}}\sqrt {\frac{3}{5}} $ and the maximum value of $f$ subject to the unit circle ${x^2} + {y^2} = 1$ is $\frac{6}{{25}}\sqrt {\frac{3}{5}} $.
