Answer
(a) using the constraint $g\left( {x,y} \right) = {\left( {x - 1} \right)^3} - {y^2} = 0$, we show that the minimum occurs when $x=1$.
Thus, the minimum occurs at $P = \left( {1,0} \right)$.
(b) using the method of Lagrange multipliers to minimize $f$ we show that the Lagrange condition is not satisfied for any value of $\lambda$.
(c) this does not contradict Theorem 1.
Work Step by Step
(a) Our task is to minimize $f\left( {x,y} \right) = x$ subject to $g\left( {x,y} \right) = {\left( {x - 1} \right)^3} - {y^2} = 0$.
We notice that ${y^2} \ge 0$. Since the right-hand side of $g$ is zero, ${\left( {x - 1} \right)^3}$ must not be negative. Thus, $x \ge 1$. Therefore, the minimum occurs when $x=1$. As a result, the constraint yields $y=0$. Hence, without using calculus, we have shown that the minimum occurs at $P = \left( {1,0} \right)$.
(b) Next, we try using the method of Lagrange multipliers to minimize $f$ subject to the constraint.
The Lagrange condition $\nabla f = \lambda \nabla g$ gives
$\left( {1,0} \right) = \lambda \left( {3{{\left( {x - 1} \right)}^2}, - 2y} \right)$
So, the Lagrange equations are
(1) ${\ \ \ \ }$ $1 = 3\lambda {\left( {x - 1} \right)^2}$, ${\ \ \ }$ $0 = - 2\lambda y$
The first equation of (1) implies that $\lambda \ne 0$. As a result, the second equation of (1) yields $y=0$ and $\lambda = \frac{1}{{3{{\left( {x - 1} \right)}^2}}}$.
Substituting $y=0$ in the constraint $g$ gives
${\left( {x - 1} \right)^3} = 0$
So, $x=1$.
Thus, the critical point is $P = \left( {1,0} \right)$. The result agrees with part (a). However, $\lambda$ is undefined at $x=1$. Therefore, we conclude that the Lagrange condition $\nabla {f_P} = \lambda \nabla {g_P}$ is not satisfied for any value of $\lambda$.
(c) In part (a) and (b) we obtain the critical point $P = \left( {1,0} \right)$. Since $\nabla g = \left( {3{{\left( {x - 1} \right)}^2}, - 2y} \right)$, the gradient $\nabla g$ at $P$ is zero, that is $\nabla {g_P} = 0$.
Recall from Theorem 1 which states that if $f\left( {x,y} \right)$ has a local minimum or a local maximum on the constraint curve $g\left( {x,y} \right) = 0$ at $P = \left( {a,b} \right)$, and if $\nabla {g_P} \ne 0$, then there is a scalar $\lambda$ such that $\nabla {f_P} = \lambda \nabla {g_P}$. However, in this exercise $\nabla {g_P} = 0$. Therefore, Lagrange condition is not applicable in this case. Hence, the result does not contradict Theorem 1.
