Answer
(a) Using Theorem 1, we show that there is a unique point $P = \left( {1,1} \right)$ on $g\left( {x,y} \right) = 1$ where $\nabla {f_P} = \lambda \nabla {g_P}$ for $\lambda = 2$.
(b) $f$ has a local maximum at $P$ subject to the constraint.
(c) Figure 14 suggests that $f\left( P \right)$ is a global maximum subject to the constraint, within the range as is shown in this figure.
Work Step by Step
(a) We have $f\left( {x,y} \right) = {x^3} + xy + {y^3}$ and the constraint $g\left( {x,y} \right) = {x^3} - xy + {y^3} = 1$.
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {3{x^2} + y,3{y^2} + x} \right) = \lambda \left( {3{x^2} - y,3{y^2} - x} \right)$
So,
(1) ${\ \ \ \ }$ $3{x^2} + y = \lambda \left( {3{x^2} - y} \right)$, ${\ \ }$ $3{y^2} + x = \lambda \left( {3{y^2} - x} \right)$
Since $\left( {0,0} \right)$ does not satisfy the constraint $g\left( {x,y} \right)=1$, we may have the following cases:
Case 1: $x=0$, $y \ne 0$
Equation (1) becomes
$y = - \lambda y$, ${\ \ \ }$ $3{y^2} = 3\lambda {y^2}$
The first equation gives $\lambda = - 1$. But $\lambda = - 1$ does not satisfy the second equation. Therefore, we conclude that there is no solution for this case.
Case 2: $x \ne 0$, $y=0$
Equation (1) becomes
$3{x^2} = 3\lambda {x^2}$, ${\ \ \ }$ $x = - \lambda x$
The first equation gives $\lambda = 1$. But $\lambda = 1$ does not satisfy the second equation. Therefore, we conclude that there is no solution for this case.
Case 3: $x \ne 0$, $y \ne 0$
From equation (1) we obtain $\lambda = \frac{{3{x^2} + y}}{{3{x^2} - y}} = \frac{{3{y^2} + x}}{{3{y^2} - x}}$.
$\left( {3{x^2} + y} \right)\left( {3{y^2} - x} \right) = \left( {3{x^2} - y} \right)\left( {3{y^2} + x} \right)$
$9{x^2}{y^2} + 3{y^3} - 3{x^3} - xy = 9{x^2}{y^2} - 3{y^3} + 3{x^3} - xy$
$6{y^3} = 6{x^3}$
$y = x$
Substituting $y = x$ in the constraint $g\left( {x,y} \right) = {x^3} - xy + {y^3} = 1$ gives
${x^3} - {x^2} + {x^3} = 1$
$2{x^3} - {x^2} - 1 = 0$
$\left( {x - 1} \right)\left( {2{x^2} + x + 1} \right) = 0$
$x - 1 = 0$ ${\ \ }$ or ${\ \ }$ $2{x^2} + x + 1 = 0$
However, the solutions for $2{x^2} + x + 1 = 0$ is
$x = \frac{{ - 1 \pm \sqrt {{1^2} - 4\cdot2\cdot1} }}{4}$,
which are complex numbers. Therefore, for real solution, there is only one solution at $x=1$.
Using $y=x$, we obtain $y=1$. Using $\lambda = \frac{{3{x^2} + y}}{{3{x^2} - y}}$, we obtain $\lambda = 2$.
Thus, the critical point is $P = \left( {1,1} \right)$.
Hence, there is a unique point $P = \left( {1,1} \right)$ on $g\left( {x,y} \right) = 1$ where $\nabla {f_P} = \lambda \nabla {g_P}$ for $\lambda = 2$.
(b) Suppose we move toward $P$ along the constraint curve $g\left( {x,y} \right) = 1$ (the red curve in Figure 14), $f$ is increasing (either we approach $P$ from the left or from the right) until we arrive at $P$, where $\nabla {f_P}$ is orthogonal to the constraint curve $g\left( {x,y} \right) = 1$. Once at $P$, we cannot increase $f$ further without leaving the constraint curve. Thus, $f$ has a local maximum at $P$ subject to the constraint.
(c) In Figure 14 we see that there is no other point $\left( {x,y} \right)$ along the constraint curve $g\left( {x,y} \right) = 1$, where the value $f\left( {x,y} \right)$ is larger than the value of $f$ at $P$. Therefore, Figure 14 suggests that $f\left( P \right)$ is a global maximum subject to the constraint, within the range as is shown in this figure.
