s. O -4 points SCalcET7 14.7.016. Notes Ask Your Find the local maximum and mini

Question

s. O -4 points SCalcET7 14.7.016. Notes Ask Your Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the Find the local maximum and minimum values and saddle pointís) of the function. I function with a domain and viewpoint that reveal all the important aspects of the function. (Enter your answers as a comma -separated list. If an answer does not exist, enter l DNE.) x, y) - e(v-x2) local maximum value(s) local minimum value(s) saddle point(s)

Explanation / Answer

f(x , y) = 2ey(y2 - x2)

critical points ==> fx = 0 , fy = 0

fx = 2ey(0 - 2x)

==> fx = -4xey

fx = 0 ==> -4xey = 0 ==> x = 0

fy = 2ey(y2 -x2) + 2ey (2y - 0)           since (uv)' = u'v + uv'

==> fy = 2ey(y2 + 2y - x2)

fy = 0 ==> 2ey(y2 + 2y - x2) = 0

==> y2 + 2y - x2 = 0   (since ey is never zero)

we have x = 0

==> y2 + 2y - 02 = 0

==> y(y + 2) = 0

==> y = 0, y= -2

Hence critical points are (0 , 0) and (0 , -2)

fxx = -4(1)ey = -4ey

fyy = 2ey(y2 + 2y - x2) + 2ey (2y + 2 - 0)

==> fyy = 2ey(y2 + 2y - x2 + 2y + 2)

==> fyy = 2ey(y2 + 4y + 2 - x2)

fxy = -4xey

D = fxxfyy - (fxy)2

at (0 , 0)

D = (-4e0)(2e0(02 + 4(0) + 2 - 02)) - (-4(0)e0)2

==> D = -4(4) - 0 = -16 < 0

as D < 0 ==> (0 , 0) is a saddle point

at (0 , 0)

D = (-4e-2)(2e-2((-2)2 + 4(-2) + 2 - 02)) - (-4(0)e-2)2

==> D = 16e-4 > 0

as D > 0 and fxx < 0 ==> (0 , -2) is local maximum

Hence local maximum at (0 , -2)

and value is 2e-2((-2)2 - 02) = 8e-2 = 1.0827

local minimum Does Not Exist : DNE

saddle point at (0 , 0)

Value is 2e0(02 - 02) = 0

==> saddlde point = (0 , 0 , 0)

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