Question

Find the points on the ellipse 2 x squared plus 3 y squared equals 1where f (x comma y )equals xyhas its extreme values.

Answer 1

(
`1/2,1/√(6)`),(-1/2,
`1/√(6)`),(-1/2,
`-1/√(6)`) and

(
`1/2,-1/√(6)`)

Explanation:

We are given that

`g(x,y)=2x^2+3y^2-1=0`

`f(x,y)=xy`

Using Lagrange's multipliers

`f_x(x,y)=y`

`f_y(x,y)=x`

`g_x(x,y)=4x`

`g_y(x,y)=6y`

`f_x(x,y)=\lambda g_x(x,y)`

`y=4x\lambda`

`f_y(x,y)=\lambda g_y(x,y)`

`x=6\lambda y`

`\lambda=(y)/(4x)`...(1)

`\lambda=(x)/(6y)`...(2)

Equation (1) divided by (2) Then we get

`1=(6y^2)/(4x^2)=(3y^2)/(2x^2)`

`2x^2=3y^2`

Substitute the values in g(x,y)

`3y^2+3y^2=1`

`6y^2=1`

`y^2=1/6`

`y=\pm (1)/(√(6))`

`2x^2+2x^2=1`

`4x^2=1`

`x^2=1/4`

`x=\pm (1)/(2)`

Therefore, the points on the ellipse are

(
`1/2,1/√(6)`),(-1/2,
`1/√(6)`),(-1/2,
`-1/√(6)`) and

(
`1/2,-1/√(6)`)