Analyzing critical points Use the Second Derivative Test to classify the critical points of f(x,y)=xy(x−2)(y+3).

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Step 1 First, We find the critical points for: f(x, y)=xy(x−2)(y+3) Derivative of equation (1), with respect to x: f(x, y)=(x2y−2xy)(y+3) f(x, y)=x2y2+3x2y−2xy2−6xy fx(x, y)=2xy2+6xy−2y2−6y Again, derivative for equation (1) with respect to y: fy(x, y)=2x2y+3x2−4xy−6x Step 2 To determine critical points: Equating equation (2) equal to 0: 2xy2+6xy−2y2−6y=0 y(2xy+6x−2y−6)=0 y=0 2xy+6x−2y−6=0 2xy+6x−2y=6 Equating equation (3) equal to 0: 2x2y+3x2−4xy−6x=0 x(2xy+3x−4y−6)=0 x=0 2xy+3x−4y−6=0 Step 3 Now, we solve the simultaneous equations (4) and (5) for values of x and y: Equation(4) can be written as: 6x−2y=6−2xy Equation(5) can be written as: 3x−4y=6−2xy We can equate (6) and (7) as: 6x−2y=3x−4y x=−23y Step 4 Put the value of x in equation (5),to get y: 2(−23y)y+3(−23y)−4y=6 −43y2−2y−4y=6 −43y2−6y=6 −43y2−6y−6=0 2y2+9y+9=0 Solving quadratic:

Analyzing critical points Use the Second Derivative Test to classify

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