# Lagrange Multiplier question with solid attempt!

## Homework Statement

Use the method of Lagrange multipliers to ﬁnd the maximum and minimum values of the function

f(x, y) = x + y2

subject to the constraint g(x,y) = 2x2 + y2 - 1

none

## The Attempt at a Solution

We need to find $\nabla$f = λ$\nabla$g

Hence,

$\nabla$fx - λ$\nabla$gx = 0

Which becomes, 1 - λ(4x) = 0

$\nabla$fy - λ$\nabla$gy = 0

Which becomes, 2y - λ = 0

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Now we have: x = 1/4λ and y = λ/2

I assume I am right in now subbing x and y into the constraint....

To give us: 2/16λ2 + λ2/4 - 1 = 0

It seems a bit messy considering this is an elementary part of my homework? Have I gone wrong somewhere?

Regards as always

Ray Vickson
Homework Helper
Dearly Missed
Your $\partial{L}/\partial{y}$ is incorrect; try again.

RGV

Your $\partial{L}/\partial{y}$ is incorrect; try again.

RGV

Wow, rookie error..

yep dL/dy = 2y - λ(2y)
so λ = 1

Thus, x=1/4

Put this into constraint to get: y=7/8

So (1/4,7/8) is a point..

How do i find the maximum and minimum? thanks

Ray Vickson
Homework Helper
Dearly Missed
the equation 2y = lambda*(2y) has TWO solutions: lambda = 1 is one of them. Can you see another?

RGV

HallsofIvy
Since specific values of $\lambda$ are not part of the "solution" of problems like these, I find it is often a good step to start by dividing one equation by another, eliminating $\lambda$ at the start. Here, your equations are $1= 4\lambda x$ and $2y= 2\lambda y[/i2tex]. Dividing the second equation by the first: $$\frac{2y}{1}= \frac{2\lambda y}{\lambda x}= \frac{2y}{x}$$ which reduces to x= 1 or y= 0. However, you say your constraint is "[itex]g(x,y)= 2x^2+ y^2- 1$" which is not a constraint at all. Did you mean $g(x,y)= 2x^2+ y^2= 1$? If so, set x= 1 in that and solve for y, then set y= 0 and solve for x.