Maximize $(x-1)2y$ subject to $x^3+y^2=3$

Question

I have to do with lagrange multiplier

f = $(x-1)2y + \lambda (x^3+y^2-3) = 0 $

I get

$f_x= y + \lambda x = 0 $

$f_y = y\lambda + x - 1 = 0 $

$x^3+y^2=3$

and

how to solve this

Answer 1

Hint.

Solving for $x, y$

$$ \left\{ \begin{array}{rcl} 3 \lambda x^2+2 y&=&0 \\ 2 (x-1)+2 \lambda y&=&0 \\ \end{array} \right. $$

we get

$$ x = \frac{1\pm\sqrt{1-6 \lambda ^2}}{3 \lambda ^2}\\ y = \frac{3\pm\frac{\sqrt{1-6 \lambda ^2}}{\lambda^2}-\frac{1}{\lambda ^2}}{3 \lambda } $$

substituting those values into $x^3+y^2 = 3$

$$ \left\{ \begin{array}{c} \frac{\left(1-\sqrt{1-6 \lambda ^2}\right)^3}{27 \lambda ^6}+\frac{\left(\frac{\sqrt{1-6 \lambda ^2}}{\lambda ^2}-\frac{1}{\lambda ^2}+3\right)^2}{9 \lambda ^2}-3=0 \\ \frac{\left(\sqrt{1-6 \lambda ^2}+1\right)^3}{27 \lambda ^6}+\frac{\left(-\frac{\sqrt{1-6 \lambda ^2}}{\lambda ^2}-\frac{1}{\lambda ^2}+3\right)^2}{9 \lambda ^2}-3=0 \\ \end{array} \right. $$

after symplifying and solving we obtain

$$ \lambda^* = \pm 0.34548248 $$

so we get

$$ \left[ \begin{array}{cc} x^*& y^*\\ 1.30482 & -0.882306 \\ 1.30482 & 0.882306 \\ 4.28061 & -9.49573 \\ 4.28061 & 9.49573 \\ \end{array} \right] $$

as the stationary points.

Answer 2

Correcting your error in computing $f_x$ the problem amounts to solution of the equation system: $$ \begin{array}{rcl} x^3+y^2&=3\\ 3 \lambda x^2+2 y&=0 \\ 2x+2 \lambda y&=2 \\ \end{array} $$

Multiplying the second equation by $\lambda$ and subtracting the third one we obtain: $$ \lambda^2=\frac{2(x-1)}{3x^2}. $$ Substituting the obtained value into the second equation one obtains: $$4y^2=9\lambda^2x^4\implies y^2=\frac32x^2(x-1),$$ which after substitution into the first equation gives rise to: $$ 5x^3-3x^2-6=0. $$ The equation has one real $$x_r=\frac{1+\omega+\omega^{-1}}5, \text{ with } \omega=\left(76+5\sqrt{231}\right)^{1/3} $$ and two complex roots. From the context of your question the complex roots can be disregarded. Since the real root is larger than $1$ the pair $$(x,y)=\left(x_r,x_r\sqrt{\frac{3(x_r-1)}2}\right)\approx(1.304821,0.882306)$$ is the only candidate for the solution (you should however check if the stationary point is indeed the global maximum).