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Lagrange multipliers: help solving for x, y and lambda

  • Thread starter rudy
  • Start date
  • #1
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6

Homework Statement


Find the local extreme values of ƒ(x, y) = x2y on the line x + y = 3

Homework Equations


∇ƒ = λ∇g

The Attempt at a Solution


2yxi+x^2j = λi + λj
[2yx=λ] [x^2=λ] [x+y=3]
[2yx=x^2] & [(2y)+y=3]
[2y=x] & [y=1]
x=2

This is as far as I got, and gives f(2,1)=4 as the extreme value. The solution I have (chegg.com) lists another possible solution.

On the left column, they give another solution to the equation 2y=x:
They say:
2y=x
x=0
and proceed to use the "x+y=3" equation to give an alternative extrema which is f(0,3)

Can anyone explain how they made the jump from 2y=x to x=0?
Why isn't y=0 another solution? (If x equals and 2y equals x, then y must equal zero)

Thanks in advance, screenshot from solution below

-Rudy

Screen Shot 2018-07-22 at 8.55.24 PM.png
 

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Answers and Replies

  • #2
tnich
Homework Helper
1,048
336

Homework Statement


Find the local extreme values of ƒ(x, y) = x2y on the line x + y = 3

Homework Equations


∇ƒ = λ∇g

The Attempt at a Solution


2yxi+x^2j = λi + λj
[2yx=λ] [x^2=λ] [x+y=3]
[2yx=x^2] & [(2y)+y=3]
[2y=x] & [y=1]
x=2

This is as far as I got, and gives f(2,1)=4 as the extreme value. The solution I have (chegg.com) lists another possible solution.

On the left column, they give another solution to the equation 2y=x:
They say:
2y=x
x=0
and proceed to use the "x+y=3" equation to give an alternative extrema which is f(0,3)

Can anyone explain how they made the jump from 2y=x to x=0?
Why isn't y=0 another solution? (If x equals and 2y equals x, then y must equal zero)

Thanks in advance, screenshot from solution below

-Rudy

View attachment 228274
You ignored the second solution when you reduced ##2xy = x^2## to ##2y=x##. Try writing it as ##2xy-x^2=0## and factoring it.
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

Homework Statement


Find the local extreme values of ƒ(x, y) = x2y on the line x + y = 3

Homework Equations


∇ƒ = λ∇g

The Attempt at a Solution


2yxi+x^2j = λi + λj
[2yx=λ] [x^2=λ] [x+y=3]
[2yx=x^2] & [(2y)+y=3]
[2y=x] & [y=1]
x=2

This is as far as I got, and gives f(2,1)=4 as the extreme value. The solution I have (chegg.com) lists another possible solution.

On the left column, they give another solution to the equation 2y=x:
They say:
2y=x
x=0
and proceed to use the "x+y=3" equation to give an alternative extrema which is f(0,3)

Can anyone explain how they made the jump from 2y=x to x=0?
Why isn't y=0 another solution? (If x equals and 2y equals x, then y must equal zero)

Thanks in advance, screenshot from solution below

-Rudy

View attachment 228274
I think their writeup was bad. It should have said "##2xy = x^2 \Rightarrow x = 0 \; {\bf \text{or}} \; 2y = x##."
 

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