Use lagrange multipliers to find the maximum and minimum values of the function f(x,y)= 4x^2 - 4xy + 4y^2 for?

Question

Use lagrange multipliers to find the maximum and minimum values of the function f(x,y)= 4x^2 - 4xy + 4y^2 for the points (x,y) that lie on the curve x^2+y^2=100. Steps please

Answer 1

Let g(x,y) = x^2+y^2=100

Find gradient vectors:

∇f = <8x-4y, -4x+8y>

∇g=<2x,2y>

Set up the equation with Lagrange multiplier:

∇f = λ∇g

That is,

8x-4y = λ(2x)---(1)

-4x+8y = λ(2y)---(2)



(1)+(2)

4x+4y=λ(2x+2y) ==> 2(x+y)=λ(x+y)

2(x+y)-λ(x+y)=0

(x+y)(2-λ) = 0

x+y= 0 or λ=2



1. x+y =0

Sub back into x^2+y^2=100

x=±√50, y=∓√50



2.λ=2

Sub back into (1) and (2)

8x-4y = 4x ==> x=y

-4x+8y = 4y ==> x=y

Sub back into x^2+y^2=100

x=±√50, y=±√50



Thus, these are the critical points:

(±√50,±√50), (±√50,∓√50)

xy = 50 or - 50

f(x,y) = 4x^2 - 4xy + 4y^2

.........= 4(x^2+y^2) - 4xy ---- x^2+y^2 = 100

.........= 400 - 4xy

Thus, the max = 400 - 4(-50) = 600

the min = 400 - 4(50) = 200

Answer 2

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Use lagrange multipliers to find the maximum and minimum values of the function f(x,y)= 4x^2 - 4xy + 4y^2 for?

Use lagrange multipliers to find the maximum and minimum values of the function f(x,y)= 4x^2 - 4xy + 4y^2 for the points (x,y) that lie on the curve x^2+y^2=100. Steps please

Answer 3

Use Lagrange Multipliers

Answer 4

f(x,y) = (xy)^1/2 + y^2 or df=[(1/2)*{(y/x)^(1/2)}]dx+ [{(1/2)*{(x/y)^(1/2)}}+2y]dy = 0 or multiplying by2* {xy)^(1/2) throughout, we get ydx +[x+4{(y)^(3/2)}*x^(1/2)]dy = 0 --------------------- (1) Also (x+y) = 3 gives dx +dy = 0 ---------------------------------(2) Adding (1) and λ times (2) we get (y+λ)dx + [x+λ +4{(y)^(3/2)}*x^(1/2)]dy = 0 So λ = -y = -x -4{(y)^(3/2)}*x^(1/2)] or = -x -4*[y*{f(x, y) - y^2] x - y = -4{(y)^(3/2)}*x^(1/2)], --------------------(3) also x+y = 3 ----------------------------------------... Adding 3 and 4 we get 2x = -4{(y)^(3/2)}*x^(1/2)] -3 or