# Help finding the largest and smallest value

• Sep 19th 2009, 12:24 PM
sebasto
[not solved]Help finding the largest and smallest value
I need to find the largest and smallest value for f(x,y) in a particular domain.

$\displaystyle f(x,y)=x^2+x(y^2-1)$ and $\displaystyle x^2+y^2\leq17$

I found the stationary points but I have trouble when I insert the domain in f(x,y) $\displaystyle g(x)=x(x+16-x^2)$ as the answer I get is not right.

How would you solve the problem?

Thanks
• Sep 19th 2009, 02:07 PM
Soroban
Hello, sebasto!

The domain gave me headaches, so I ignored it at the beginning.

Quote:

Find the maximum and minimum for $\displaystyle f(x,y)$ in a given domain.

. . $\displaystyle f(x,y)\:=\:x^2+xy^2-x \qquad\text{Domain: }x^2+y^2\:\leq\:17$

Equate the partial derivative to zero and solve.

. . $\displaystyle \begin{array}{ccccc}f_x\!: & 2x + y^2 - 1 \:=\: 0 & [1] \\ f_y\!: & 2xy \:=\: 0 & [2] \end{array}$

From [2], we have: .$\displaystyle x = 0,\;y = 0$

If $\displaystyle x = 0$, [1] becomes: .$\displaystyle y^2 - 1 \:=\:0 \quad\Rightarrow\quad y \:=\:\pm1$
. . We have two critical points: .$\displaystyle (0,\:\pm1)$

If $\displaystyle y = 0$, [1] becomes: .$\displaystyle 2x - 1 \:=\:0 \quad\Rightarrow\quad x \:=\:\tfrac{1}{2}$
. . We have one more critical point: .$\displaystyle \left(\tfrac{1}{2},\:0\right)$

Second Partial Test: .$\displaystyle \begin{Bmatrix}f_{xx} &=& 2 \\ f_{yy} &=& 2x \\ f_{xy} &=& 2y \end{Bmatrix} \quad\Rightarrow\quad D \;=\;4x- 4y^2$

At $\displaystyle (0,\:\pm1)\!:\;\;D \:=\:4(0)- 4(\pm1)^2 \:=\:-4$ . . . saddle points

At $\displaystyle \left(\tfrac{1}{2},\:0\right)\!:\;\;D \:=\:4\left(\tfrac{1}{2}\right) - 4(0^2) \:=\:+2$ . . . extremum

. . $\displaystyle f_{xx} = +2$ . . . positive, concave up

Therefore: .$\displaystyle \text{ minimum at }\left(\frac{1}{2},\:0,\:\text{-}\frac{1}{4}\right)$ . . . which is in the domain.

Edit: We must find the maximum(s) along the circumference of that circle.

I would use Lagrange multipliers: .$\displaystyle F(x,y,\lambda) \;=\;x^2 + xy^2 - x + \lambda(x^2+y^2-17)$

• Sep 19th 2009, 02:43 PM
sebasto
I got the same saddle points and minimum with their respective function value, but the answer is

smallest value $\displaystyle f(-2;\sqrt{13})=-20$
Largest value $\displaystyle f(-\sqrt{17};0)=17+\sqrt{17}$

Maybe I formulated the question wrong, I'm looking for the smallest and largest value in the range $\displaystyle S=\{(x,y):x^2+y^2\leq17\}$
• Sep 20th 2009, 05:25 AM
sebasto
Anyone out there that has a clue on how to solve this? I have a test tomorrow and would really appreciate knowing how to solve this problem.

Thanks (Hi)