What are the values and types of the critical points, if any, of #f(x, y) = x^3+y^3-3*x*y-7#?

Question

Isabella Ackerman · Accepted Answer

There is a saddle point at #(0,0)# and a relative minimum at #(1,1)#

Here, we have

#f(x,y)=x^3+y^3-3xy-7#

Compute the partial derivatives

#f_x=3x^2-3y#

#f_y=3y^2-3x#

#f_x=0#, #=>#, #x^2=y#

#f_y=0#, #=>#, #y^2=x#

#x^4=y^2=x#

#x^4-x=0#

#x(x^3-1))=0#

#x(x-1)(x^2+x+1)=0#

#x=0# and #x=1#

#y=0# and #y=1#

The critical point are #(0,0)# and #(1,1)#

Now, compute the second partial derivatives

#f_(x x)=6x#

#f_(yy)=6y#

#f_(xy)=-3#

#f_(yx)=-3#

Therefore,

#D(x,y)=f_(x x) f_(yy)-f_(xy)^2=(6x)(6y)-(-3)^2=36xy-9#

#D(0,0)=-9#

This point #(0,0)# corresponds to a saddle point

#D(1,1)=36-9)=27# and #f_(x x)(1,1)=6#

There is a relative min at #(1,1)#

William Abdalla · Answer

undefined To find the critical points of $ f(x, y) = x^3 + y^3 - 3xy - 7 $, we first compute the partial derivatives with respect to $ x $ and $ y $, and then set them equal to zero and solve for $ x $ and $ y $. The partial derivatives are:

$$ \frac{\partial f}{\partial x} = 3x^2 - 3y $$
$$ \frac{\partial f}{\partial y} = 3y^2 - 3x $$

Setting these equal to zero and solving for $ x $ and $ y $, we find:

$$ 3x^2 - 3y = 0 \implies x^2 = y $$
$$ 3y^2 - 3x = 0 \implies y^2 = x $$

Solving these simultaneous equations yields critical points at $ (1,1) $ and $ (-1,-1) $. To determine the nature of these critical points, we evaluate the second partial derivatives and use the second derivative test.

The second partial derivatives are:

$$ \frac{\partial^2 f}{\partial x^2} = 6x $$
$$ \frac{\partial^2 f}{\partial y^2} = 6y $$
$$ \frac{\partial^2 f}{\partial x \partial y} = -3 $$

At the critical point $ (1,1) $, the second partial derivatives are all positive, indicating that $ f(x, y) $ has a local minimum at $ (1,1) $. Similarly, at the critical point $ (-1,-1) $, the second partial derivatives are all positive, indicating a local minimum as well.

Therefore, the values and types of the critical points of $ f(x, y) = x^3 + y^3 - 3xy - 7 $ are:

1. $ (1,1) $: Local minimum
2. $ (-1,-1) $: Local minimum