Find the maximum and minimum values of http://webwork.math.ucdavis.edu/webw...34de4622b1.png on the disk D: http://webwork.math.ucdavis.edu/webw...8ec2ef0161.png.

maximum value:

minimum value:

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- Apr 28th 2010, 03:39 PMewkimchiFind the maximum and minimum values of
Find the maximum and minimum values of http://webwork.math.ucdavis.edu/webw...34de4622b1.png on the disk D: http://webwork.math.ucdavis.edu/webw...8ec2ef0161.png.

maximum value:

minimum value: - Apr 29th 2010, 01:25 PMhollywood
Well, you can probably see by inspection that (x,y)=(0,0) is a minimum, but let's look at the critical points - where $\displaystyle \frac{\partial{f}}{\partial{x}}=\frac{\partial{f}} {\partial{y}}=0$.

$\displaystyle \frac{\partial{f}}{\partial{x}}=12x$

$\displaystyle \frac{\partial{f}}{\partial{y}}=14y$

so the only critical point is (0,0). But we have to look at the boundary, that is, where $\displaystyle x^2+y^2=1$. Solving for $\displaystyle y^2$ and substituting gives:

$\displaystyle 6x^2+7y^2=6x^2+7(1-x^2)=7-x^2$, which has a maximum at x=0, and minima at x=1 and -1. The minima are greater than the minimum at (0,0), so we can ignore them. But the maximum at x=0 gives us maxima at (x,y)=(0,1) and (0,-1).

- Hollywood