Multi Max
Calculus Max/Min Partial Derivative Help!?
I have this assignment and the final question has me stumped. I’ve tried getting some help but it didn’t really help. Hoping Yahoo Answers could help me out. The question is:
Max/min of a multi-variable function over a closed and bounded region.
Find the absolute maximum and minimum values of:
f (x,y) = e ^ (x^2 + 2y^2) over the disk D given by x^2 + y^2 (smaller or equal to ) 4
If anyone knows how to come to the solution, if would be greatly appreciated. Thanks in advance!
The constraint is x^2 + y^2 <= 4 --> y^2 <= 4 - x^2 or 2y^2 <= 8 - 2x^2.
e^{x^2 + 2y^2} <= e^{x^2 + 8 - 2x^2} = e^{8 - x^2} which we can call g(x)
g'(x) = e^(8 - x^2) * (-2x) dx.
Setting g'(x) = 0 will find relative extrema, but g'(x) will = 0 only when x = 0 since e^(8-x^2) will never = 0.
Back to the constraint formula. If x = 0, then the constraint is y^2 <= 4 --> y <= |2|.
Obviously, y will attain a minimum value at y = 0 and a maximum value at y = +/- 2.
When y = 0, e^(x^2 + 2y^2) = e^(0 + 0) = 1.
When y = -2 or 2, e^(x^2 + 2y^2) = e^(0 + 
= 2980.957987.
If x = y = sqrt(2) (which satisfies the constraint), then e^(x^2 + 2y^2) = e^6 = 403.4287935.
If x = 1 then y = sqrt(3) and e(x^2 + 2y^2) = e^7 = 1096.633158.
In short, the absolute minimum occurs at (0,0) and the absolute maximum occurs at (0,-2) and (0,2).
Chage & Multi-Max with: Someday
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