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History Bookmarks Window Help webassign.net This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagra find the extreme values of the function subject to the given constraint fx, y, z) xy2z; xy2+2 36 maximum value 216 minimum value2 Need Help?adTalk to a Tutor 216 x My 4. 0/2 points| Previous Answers SCalcET8 14.8.509.XPMI Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given answer does not exist, enter DNE) fx, y)- xy x2 +2y2-6 maximum DNE minimum DNE My This question has several parts that must be completed sequentially. If you skip a part of the question, you any points for the skipped part, and you will not be able to come back to the skipped part 5. 8/14 points 1 Previous Answers SCaldET8 14.8.509.XP.MI.SA Use Lagrange multipliers to find the maximum and minimum values of the function subject to the constraint. 5

Explanation / Answer

3)

using lagrange multiplier

<y²z , 2xyz, xy²> =k <2x, 2y, 2z>

Comparing both sides

k= y²z/(2x) =2xyz/(2y) =xy²/(2z)

Hence

z² =x²

y²=2x²

substitute this in constarint eqaution x²+y²+z²=36

x² +2x² +x²= 36

4x² =36

x² =9

x= -3, 3

hence

y= -3*sqrt(2) , 3*sqrt(2)

z = -3, 3

Now

f(3, 3*sqrt(2), 3) = 162

f(3, 3*sqrt(2), -3) = -162

hence

maximum value = 162

Minimum value = -162

=============

4)

maximum value = 4

Minimum value = -4

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