Answer:
([tex]1/2,1/\sqrt{6}[/tex]),(-1/2,[tex]1/\sqrt{6}[/tex]),(-1/2,[tex]-1/\sqrt{6}[/tex]) and
([tex]1/2,-1/\sqrt{6}[/tex])
Step-by-step explanation:
We are given that
[tex]g(x,y)=2x^2+3y^2-1=0[/tex]
[tex]f(x,y)=xy[/tex]
Using Lagrange's multipliers
[tex]f_x(x,y)=y[/tex]
[tex]f_y(x,y)=x[/tex]
[tex]g_x(x,y)=4x[/tex]
[tex]g_y(x,y)=6y[/tex]
[tex]f_x(x,y)=\lambda g_x(x,y)[/tex]
[tex]y=4x\lambda[/tex]
[tex]f_y(x,y)=\lambda g_y(x,y)[/tex]
[tex]x=6\lambda y[/tex]
[tex]\lambda=\frac{y}{4x}[/tex]...(1)
[tex]\lambda=\frac{x}{6y}[/tex]...(2)
Equation (1) divided by (2) Then we get
[tex]1=\frac{6y^2}{4x^2}=\frac{3y^2}{2x^2}[/tex]
[tex]2x^2=3y^2[/tex]
Substitute the values in g(x,y)
[tex]3y^2+3y^2=1[/tex]
[tex]6y^2=1[/tex]
[tex]y^2=1/6[/tex]
[tex]y=\pm \frac{1}{\sqrt{6}}[/tex]
[tex]2x^2+2x^2=1[/tex]
[tex]4x^2=1[/tex]
[tex]x^2=1/4[/tex]
[tex]x=\pm \frac{1}{2}[/tex]
Therefore, the points on the ellipse are
([tex]1/2,1/\sqrt{6}[/tex]),(-1/2,[tex]1/\sqrt{6}[/tex]),(-1/2,[tex]-1/\sqrt{6}[/tex]) and
([tex]1/2,-1/\sqrt{6}[/tex])
