Maximum and Minimum values of a function (2 vars)

  • Thread starter VenaCava
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  • #1
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Homework Statement



Find the max and min values of the function f(x, y) = (x-y)/(2 +x2 + y2)
over the disk x2 + y2 <= 4

Homework Equations




The Attempt at a Solution




fx = 1/(2 +x2 + y2) + -2x(x-y)/(2 +x2 + y2)2
= (2- x2 + y2 +2xy)/(2 +x2 + y2)2

Should equal zero when x=1 y= -1 or x=-1 y=1

fy= (-2- x2 + y2 -2xy)/(2 +x2 + y2)2

equals zero when x=1 y=-1 or y=1 x=-1

Critical points at (1, -1) and(-1, 1)

f(-1, 1)= -1/2 f(1, -1) = 1/2

If I check the boundary

I get f(x, (4-x2)1/2) =1/6( x - (4-x2)1/2) _

If I take the derivative of that and set it to zero, I don't see any values that will let it equal 0.

f' = 1/6(1 + x/(4-x2)1/2

Shouldn't there be a max and min value on the boundary, even if it's not the abs max/min on the region.

I know I must be making a mistake somewhere, but I can't find it.

Thanks
 

Answers and Replies

  • #2
It seems probable that your forgetting a rule that happens when you take square roots of a value. (I do it all the time. Drives me nuts, so if this is your error I understand your pain.)

As just a random example, If x2 = 9, how many different values get you to the 9?
 

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