Extrema with constraints

Mar 2009
77
2
Here is the question:

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

My attempts at solution:

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

\(\displaystyle 1 = 4x \lambda \) ............................. I

\(\displaystyle 2y = 2y \lambda\) ............................. II

\(\displaystyle 2x^2 + y^2 = 1 \) ............................. 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?
 
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May 2010
274
67
Los Angeles, California
Doesn't equation II tell you that \(\displaystyle \lambda =1\) (assuming \(\displaystyle y\ne 0 )\)?
 
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Jul 2007
894
298
New Orleans
Here is the question:

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

My attempts at solution:

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

\(\displaystyle 1 = 4x lambda \) ............................. I

\(\displaystyle 2y = 2y lambda\) ............................. II

\(\displaystyle 2x^2 + y^2 = 1 \) ............................. 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?

\(\displaystyle 2y-2y\lambda = 0\)

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

\(\displaystyle y = 0, \lambda = 1, x = \pm \frac{1}{\sqrt{2}}\)
 
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May 2010
20
8
Your system of equations is correct.

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

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


SO, your 4 critical points are...
\(\displaystyle x=\frac{1}{4}\) , \(\displaystyle y=\pm\frac{1}{2}\sqrt\frac{7}{2}\)
\(\displaystyle y=0\) , \(\displaystyle x=\pm\frac{1}{\sqrt2}\)
 

matheagle

MHF Hall of Honor
Feb 2009
2,763
1,146
I'd use Calc 1.

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

\(\displaystyle f(x)=x+1-2x^2\)
 
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Mar 2009
77
2
Doesn't equation II tell you that \(\displaystyle \lambda =1\) (assuming \(\displaystyle y\ne 0 )\)?
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 \(\displaystyle \lambda \) is 1. Substitute that in equation 1 and we get x as 1/4. Substitute for x in equation III we get y = +/- \(\displaystyle \sqrt {7/8}\)
 
Jul 2007
894
298
New Orleans
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 \(\displaystyle \lambda \) is 1. Substitute that in equation 1 and we get x as 1/4. Substitute for x in equation III we get y = +/- \(\displaystyle \sqrt {7/8}\)
Sorry forgot to give you your other x point to try
\(\displaystyle x = \frac{1}{4}\)
 
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Mar 2009
77
2
So the maximum is \(\displaystyle x=\frac{1}{4}\) , \(\displaystyle y=\sqrt \frac{7}{8}\) and the minimum is \(\displaystyle x=\frac{-1}{\sqrt2}, y=0 \)