Answer to Question #300305 in Calculus for sha

Question #300305

Locate and classify all critical points for f(x, y) = 4y^2x + 3xy


1
Expert's answer
2022-02-21T16:12:26-0500

Find the critical point(s)


fx=4y2+3y\dfrac{\partial f}{\partial x}=4y^2+3y

fy=8xy+3x\dfrac{\partial f}{\partial y}=8xy+3x

Then


{fx=0fy=0\begin{cases} \dfrac{\partial f}{\partial x}=0 \\ \\ \dfrac{\partial f}{\partial y}=0 \end{cases}

{4y2+3y=08xy+3x=0\begin{cases} 4y^2+3y=0 \\ 8xy+3x=0 \end{cases}

Critical points (0,0),(0,3/4)(0, 0), (0, -3/4)

Point (0,0)(0,0)


2fx2(0,0)=0\dfrac{\partial^2 f}{\partial x^2}(0,0)=0

2fy2(0,0)=8(0)=0\dfrac{\partial^2 f}{\partial y^2}(0,0)=8(0)=0

2fxy(0,0)=2fyx(0,0)=8(0)+3=3\dfrac{\partial^2 f}{\partial x\partial y}(0,0)=\dfrac{\partial^2 f}{\partial y\partial x}(0,0)=8(0)+3=3


D(0,0)=0330=9<0D(0,0)=\begin{vmatrix} 0 & 3 \\ 3 & 0 \end{vmatrix}=-9<0

Then the f(0,0)f(0,0) is not a local maximum or minimum.

The critical point (0,0)(0,0) is a saddle point.


Point (0,3/4)(0,-3/4)


2fx2(0,3/4)=0\dfrac{\partial^2 f}{\partial x^2}(0,-3/4)=0

2fy2(0,3/4)=8(0)=0\dfrac{\partial^2 f}{\partial y^2}(0,-3/4)=8(0)=0

2fxy(0,3/4)=2fyx(0,3/4)\dfrac{\partial^2 f}{\partial x\partial y}(0,-3/4)=\dfrac{\partial^2 f}{\partial y\partial x}(0,-3/4)

=8(3/4)+3=3=8(-3/4)+3=-3


D(0,0)=0330=9<0D(0,0)=\begin{vmatrix} 0 & -3 \\ -3 & 0 \end{vmatrix}=-9<0

Then the f(0,3/4)f(0,-3/4) is not a local maximum or minimum.

The critical point (0,3/4)(0,-3/4) is a saddle point.



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