Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find the max and min values of the function $f(x,y)=x^2+y^2$ under the restriction $g(x,y)=\frac{x^2}{2}+y^2-1=0 $

Note that we can use Lagrange Multipliers Theorem , since $grad(g(p))\ne 0$ $\forall p$ in the ellipse. Then we put the system of equations $$ \left( {2x,2y} \right) = \nabla f\left( {x,y} \right) = \lambda \nabla g\left( {x,y} \right) = \left( {\lambda x,2\lambda y} \right) $$

Then if $x,y \ne 0 $ I have two values for $ \lambda =1,2$ But I can't find $x,y$. Maybe I'm doing something wrong. Please help me.

share|improve this question
    
You can also substitute $y^2 = - \frac{x^2}{2} + 1$, and find the max and min values of $\frac{x^2}{2} + 1$ on the interval $-\sqrt{2} \leq x \leq \sqrt{2}$. $\frac{x^2}{2} + 1$ is a parabola so we don't need calculus. $x = 0$ is a minimizer (min value is $1$) and $x = \pm \sqrt{2}$ are maximizers (max value is $2$). –  littleO Jul 9 '13 at 7:52

1 Answer 1

up vote 0 down vote accepted

You want to solve the system $$ \frac{x^2}{2}+y^2-1=0\\ 2x=\lambda x\\ 2y=2\lambda y $$ for $(x,y,\lambda)$. Clearly any triple with both $x$ and $y$ non-zero is not a solution because it would force $\lambda$ to be equal to both $1$ and $2$. Similarly, any triples with both $x$ and $y$ equal to zero are not solutions because such triples cannot satisfy the first equation.

If $x=0$ and $y\neq 0$, your system of equations reduces to $$ y^2=1,\quad 2y=2\lambda y, $$ with solutions $(0,\pm1,1)$. These two points correspond to values of your target function $x^2+y^2$ of $1$.

If $y=0$ and $x\neq 0$, the system of equations simplifies to $$ x^2=2,\quad 2x=\lambda x, $$ with solutions $(\pm\sqrt{2},0,1)$; the value of the target function in this case is $2$.

The maximum of the function $(x,y)\mapsto x^2+y^2$ under the constraint $x^2/2+y^2=1$ is thus equal to $2$, and the minimum is equal to $1$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.