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How did he get the derivative? fx (x, y)=-2xety–x²-y² far (x, y)=-2e4v–x²-y² + 4x²e4y=x²-y² S ー- 2Vye fyy (x, y)=-8ye+=x²-y° – 2e+y¬x²¬y² + 4y²e4y¬x²¬y² | fxy (x, y)=-2x (4 – 2y)e4y-x²-y² Now, find the stationary points. -2retyーx-y=0 x=0 : eーード +0| (4 – 2y)e4y¬x²-y° =0 y=2

How did he get the derivative? fx (x, y)=-2xety–x²-y² far (x, y)=-2e4v–x²-y² + 4x²e4y=x²-y² S ー- 2Vye fyy (x, y)=-8ye+=x²-y° – 2e+y¬x²¬y² + 4y²e4y¬x²¬y² | fxy (x, y)=-2x (4 – 2y)e4y-x²-y² Now, find the stationary points. -2retyーx-y=0 x=0 : eーード +0| (4 – 2y)e4y¬x²-y° =0 y=2

College Algebra (MindTap Course List)
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter2: Functions And Graphs
Section2.6: Proportion And Variation
Problem 22E: Find the constant of proportionality. z is directly proportional to the sum of x and y. If x=2 and...
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How did he get the derivative? fx (x, y)=-2xey–x²-y² far (x, y)=-2e4y–x²-y² + 4x?e4y¬x²-y² ーは- 2ye fyy (x, y)=-8ye4y=x²-y° – 2e4y¬x²-y? + 4y²e+y¬x²¬y² fxy (X, y)=-2x (4 – 2y)e4y-x²-y² Now, find the stationary points. -2xeyーxーy=0 x=0 [:: e+¬x²-y° + 0] (4 – 2y)e4y¬x²-y° =0 y=2
Transcribed Image Text:How did he get the derivative? fx (x, y)=-2xey–x²-y² far (x, y)=-2e4y–x²-y² + 4x?e4y¬x²-y² ーは- 2ye fyy (x, y)=-8ye4y=x²-y° – 2e4y¬x²-y? + 4y²e+y¬x²¬y² fxy (X, y)=-2x (4 – 2y)e4y-x²-y² Now, find the stationary points. -2xeyーxーy=0 x=0 [:: e+¬x²-y° + 0] (4 – 2y)e4y¬x²-y° =0 y=2
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