Find extrema of the function z = 3x ^ 2 +2 xy +2 y ^ 2 with respect to the constraint equation 4x ^ 2 +2 y ^ 2 = 9.Write the Lagrangian,make necessary conditions,find the critical points of the...

Find extrema of the function z = 3x ^ 2 +2 xy +2 y ^ 2 with respect to the constraint equation 4x ^ 2 +2 y ^ 2 = 9.

Write the Lagrangian,make necessary conditions,find the critical points of the Lagrangian.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to determine the first partial derivatives, such that:

`f_x = (del(3x^2 + 2xy + 2y^2))/(del x) => f_x = 6x + 2y`

`f_y = (del(3x^2 + 2xy + 2y^2))/(del y) =>f_y = 2x + 4y`

You need to notice that `f_x = 0` if `6x = -2y => 3x = -y` and f_y = 0 if `x = -2y` such that:

`{(x = -y/3),(x = -2y):} => -y/3 = -2y => y = 0 => x = 0.`

Hence, the condition `f_x=0` and `f_y = 0` is satisfied at the point `(0,0)` , hence `(0,0)` is a stationary point.

You need to solve the boundary problem, using Lagrange multipliers, such that:

`g(x,y) = 4x^2 + 2y^2 => grad g(x,y) = <8x ; 8y>`

Using Lagrange, yields:

`grad f(x,y) = lambda grad g(x,y) => {(6x + 2y = lambda(8x)),(2x + 4y = lambda(8y)),(4x^2 + 2y^2 = 9):}`

You need to use the second equation and you need to divide by 2 both sides such that:

`x + 2y = 4lambda y => x = 4lambda y - 2y`

You need to substitute `4lambda y - 2y` for `x` in the first and the third equations, such that:

`3(4lambda y - 2y) + y = 4lambda(4lambda y - 2y)`

`12 lambda y - 6y + y = 16lambda^2 y - 8 lambda y`

`20 lambda y - 5y = 16lambda^2 y => 5y(4lambda - 1) = 16lambda^2 y => 5y(4lambda - 1) - 16lambda^2 y = 0`

Factoring out y yields:

`y(20 lambda - 5 - 16 lambda^2) = 0`

Hence, either `y = 0` , or `-16lambda^2 + 20 lambda - 5 = 0` .

If `y = 0 => 4x^2 = 9 => x_(1,2) = +-sqrt(9/4) => x_(1,2) = +-3/2`

Hence, you have obtained two points `(3/2,0)` and `(-3/2,0)` .

Assuming that `-16lambda^2 + 20 lambda - 5 = 0` yields:

`16lambda^2 - 20 lambda + 5 = 0`

Using quadratic formula yields:

`lambda_(1,2) = (20 +- sqrt(400 - 320))/32 => lambda_(1,2) = (20 +- 4sqrt5)/32 => lambda_(1,2) = (5 +- sqrt5)/8`

Substituting `(5 + sqrt5)/8` for `lambda` in the first equation yields:

`x = 4lambda y - 2y => x = 2y(2lambda - 1) => x = 2y((5 + sqrt5)/4 - 1) => x = 2y(1+sqrt5)/4 => x = y(1+sqrt5)/2`

Substituting `y(1+sqrt5)/2` for x in the third equation yields:

`4x^2 + 2y^2 = 9 => 4y^2(1+sqrt5)^2/4 + 2y^2 = 9`

`y^2(1 + 5 + 2sqrt5 + 2) = 9 => y_(1,2) = +-3/(sqrt(8+2sqrt5))`

`x_(1,2) = +-(3(1+sqrt5))/(2sqrt(8+2sqrt5))`

Substituting `(5 - sqrt5)/8` for `lambda` in the first equation yields:

`x = 4lambda y - 2y => x = 2y(2lambda - 1) => x = 2y((5 - sqrt5)/4 - 1) => x = 2y(1-+sqrt5)/4 => x = y(1-sqrt5)/2`

`y_(3,4) = +-3/(sqrt(8-2sqrt5)) => x_(3,4) = +-(3(1-sqrt5))/(2sqrt(8-2sqrt5))`

Hence, evaluating the critical points, using Lagrange multipliers, yields `(3/2,0)` and `(-3/2,0)` , `((3(1+sqrt5))/(2sqrt(8+2sqrt5)), 3/(sqrt(8+2sqrt5)))` and `((-3(1+sqrt5))/(2sqrt(8+2sqrt5))` ,` -3/(sqrt(8+2sqrt5))), ` `((3(1-+sqrt5))/(2sqrt(8-2sqrt5)), 3/(sqrt(8-2sqrt5)))` and `((-3(1-sqrt5))/(2sqrt(8-2sqrt5)), -3/(sqrt(8-2sqrt5)))` and you may decide what are the maximum or minimum points, evaluating the function at these points.

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