# Extrema with constraints

• May 19th 2010, 12:28 PM
Keep
Extrema with constraints
Here is the question:

Find the extrema of $f(x, y) = x + y^2$, with constraint : $2x^2 + y^2 = 1$

My attempts at solution:

I tried to solve it using Lagrange multiplier and I got:

$1 = 4x \lambda$ ............................. I

$2y = 2y \lambda$ ............................. II

$2x^2 + y^2 = 1$ ............................. III

Now here is where I am stuck. How can I use the first two equations to get the value of either x or y?
• May 19th 2010, 03:04 PM
ojones
Doesn't equation II tell you that $\lambda =1$ (assuming $y\ne 0 )$?
• May 19th 2010, 03:11 PM
11rdc11
Quote:

Originally Posted by Keep
Here is the question:

Find the extrema of $f(x, y) = x + y^2$, with constraint : $2x^2 + y^2 = 1$

My attempts at solution:

I tried to solve it using Lagrange multiplier and I got:

$1 = 4x lambda$ ............................. I

$2y = 2y lambda$ ............................. II

$2x^2 + y^2 = 1$ ............................. III

Now here is where I am stuck. How can I use the first two equations to get the value of either x or y?

$2y-2y\lambda = 0$

$2y(1 - \lambda) = 0$

$y = 0, \lambda = 1, x = \pm \frac{1}{\sqrt{2}}$
• May 19th 2010, 03:13 PM
DrDank
Your system of equations is correct.

From eq. II, $\lambda=1$
Plug that into eq. I, $x=\frac{1}{4}$
Plug $x=\frac{1}{4}$ into eq. III, $y=\pm\frac{1}{2}\sqrt\frac{7}{2}$

The other critical point occurs when $y=0$,
Use eq.III , $x=\frac{1}{\sqrt2}$

SO, your 4 critical points are...
$x=\frac{1}{4}$ , $y=\pm\frac{1}{2}\sqrt\frac{7}{2}$
$y=0$ , $x=\pm\frac{1}{\sqrt2}$
• May 19th 2010, 03:15 PM
matheagle
I'd use Calc 1.

Since $y^2=1-2x^2$ you only need to get the max's and min's of

$f(x)=x+1-2x^2$
• May 19th 2010, 03:17 PM
Keep
Quote:

Originally Posted by ojones
Doesn't equation II tell you that $\lambda =1$ (assuming $y\ne 0 )$?

Ooh OK I didn't look at it like that, I was trying to divide equation 1 by 2 and it wan't working. Yeah equation 2 says $\lambda$ is 1. Substitute that in equation 1 and we get x as 1/4. Substitute for x in equation III we get y = +/- $\sqrt {7/8}$
• May 19th 2010, 03:24 PM
11rdc11
Quote:

Originally Posted by Keep
Ooh OK I didn't look at it like that, I was trying to divide equation 1 by 2 and it wan't working. Yeah equation 2 says $\lambda$ is 1. Substitute that in equation 1 and we get x as 1/4. Substitute for x in equation III we get y = +/- $\sqrt {7/8}$

Sorry forgot to give you your other x point to try
$x = \frac{1}{4}$
• May 20th 2010, 01:11 AM
Keep
So the maximum is $x=\frac{1}{4}$ , $y=\sqrt \frac{7}{8}$ and the minimum is $x=\frac{-1}{\sqrt2}, y=0$