# find extreeemes

• Sep 11th 2012, 11:09 PM
jacob93
find extreeemes
$\displaystyle f(x,y)=x+xy^2$
only def for $\displaystyle x^2+y^2 \leq 4$
find the highest and lowest value for $\displaystyle f$.

Attempt: https://dl.dropbox.com/u/586465/Phot...2 09 04 42.jpg
I find (0,0) and (±√4 ; ±2/3*√6)

The book says the answer is: Skärmavbild 2012-09-12 kl. 09.08.23.png
• Sep 12th 2012, 12:21 AM
kalyanram
Re: find extreeemes
In your post g'(x) calculation is incorrect. In case you want to verify see the solution below.
Spoiler:

Quote:

Originally Posted by jacob93
$\displaystyle f(x,y)=x+xy^2$
only def for $\displaystyle x^2+y^2 \leq 4$
find the highest and lowest value for $\displaystyle f$.

Consider
$\displaystyle f(x,y)=x+xy^2 = x(1+y^2)$ $\displaystyle f(x,y) \ge 0 x \ge 0$ and $\displaystyle f(x,y) < 0 x < 0$ one conclusion that can be drawn is $\displaystyle f(x,y)$ reaches its extremes under the restriction $\displaystyle x^2+y^2 \leq 4$ at precisely on the circle $\displaystyle x^2+y^2 = 4$. So substitute $\displaystyle y^2 = 4 -x^2$ in $\displaystyle f(x,y)$. We have $\displaystyle g(x) = x(5-x^2)$ Equating the derivative to zero we have $\displaystyle g'(x) = 5- 3x^2 = 0 \implies x = \pm \sqrt{\frac{5}{3}}$
• Sep 13th 2012, 12:53 AM
jacob93
Re: find extreeemes
thanks