Identify the Critical Points for the function? : # f(x,y) = 3xy-x^3-3y^2 #

2 Answers
Aug 3, 2017

Answer:

The saddle point is #(0,0)# and the local maximum is #(1/2,1/4)#

Explanation:

Our function is

#f(x,y)=3xy-x^3-3y^2#

The partial derivatives are

#f_x(x,y)=3y-3x^2#

#f_y(x,y)=3x-6y#

#f_(x x)(x,y)=-6x#

#f_(y y)(x,y)=-6#

#f_(x y)(x,y)=3#

#f_(y x)(x,y)=3#

We look for the critical points

#f_x(x,y)=3y-3x^2=0#, #=>#, #y=x^2#...........#(1)#

#f_y(x,y)=3x-6y=0#, #=>#, #y=x/2#.................#(2)#

We solve for #x# and #y# in equations #(1)# and #(2)#

#x^2=x/2#

#x^2-x/2=0#

#x(x-1/2)=0#

#x=0# and #x=1/2#

Therefore, the critical points are

#(0,0)# and #(1/2,1/4)#

Therefore,

#f_(x x)(0,0)=-6*0=0# and #f_(x x)(1/2,1/4)=-6*1/2=-3#

#f_(y y)(0,0)=-6# and #f_(y y)(1/2,1/4)=-6#

#f_(x y)(0,0)=3# and #f_(x y)(1/2,1/4)=3#

To perform the second derivative test, we calculate the determinant,

#D(0,0) = | (0,3) , (3,-6) | # #= (0) * (-6)-(3) * (3) =-9 <0#, this represents a saddle point

#D(1/2,1/4) = | (-3,3) , (3,-6) | # #= (-3) * (-6)-(3) * (3) =18-9=9>0#, this represents a local maximum as #f_(x x)(1/2,1/4) <0#

Aug 3, 2017

Answer:

# (0,0) #- Saddle Point

# (1/2,1/4) # - Maximum

Explanation:

We have:

# f(x,y) = 3xy-x^3-3y^2 #

Step 1 - Find the Partial Derivatives

We compute the partial derivative of a function of two or more variables by differentiating wrt one variable, whilst the other variables are treated as constant. Thus:

The First Derivatives are:

# f_x \ = (partial f) / (partial x) \ \ = 3y -3x^2 #
# f_y \ = (partial f) / (partial y) \ \ = 3x-6y #

The Second Derivatives are:

# f_(x x) =(partial^2 f) / (partial x^2) = -6x #
# f_(yy) = (partial^2 f) / (partial y^2) = -6 #

The Second Partial Cross-Derivatives are:

# f_(xy) =(partial^2 f) / (partial x partial y) =3 #
# f_(yx) = (partial^2 f) / (partial y partial x) =3 #

Note that the second partial cross derivatives are identical due to the continuity of #f(x,y)#.

Step 2 - Identify Critical Points

A critical point occurs at a simultaneous solution of

# f_x = f_y = 0 iff (partial f) / (partial x) = (partial f) / (partial y) = 0#

i.e, when:

# 3y -3x^2 = 0# ..... [A]
# 3x-6y \ \ = 0 # ..... [B]

From [A] we have: #y=x^2# so substituting for #y# in [B] we get:

# 3x-6x^2 = 0 => 2x^2-x =0 #
# :. x(2x-1) = 0 #
# :. x=0,1/2 #

And the using #y=x^2# we have:

# x=0 \ \ => y = 0 #
# x=1/2 => y = 1/4 #

So we can conclude that there are two critical points:

# (0,0) #; and # (1/2,1/4) #

Step 3 - Classify the critical points

In order to classify the critical points we perform a test similar to that of one variable calculus using the second partial derivatives and the Hessian Matrix.

# Delta = H f(x,y) = | ( f_(x x) \ \ f_(xy) ) , (f_(yx) \ \ f_(yy)) | = | ((partial^2 f) / (partial x^2),(partial^2 f) / (partial x partial y)), ((partial^2 f) / (partial y partial x), (partial^2 f) / (partial y^2)) | = f_(x x)f_(yy)-(f_(xy))^2 #

Then depending upon the value of #Delta#:

# {: (Delta>0, "There is maximum if " f_(x x)<0),(, "and a minimum if " f_(x x)>0), (Delta<0, "there is a saddle point"), (Delta=0, "Further analysis is necessary") :} #

Using custom excel macros the function values along with the partial derivative values are computed as follows:

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