Question

Examine the function f(x,y) = x^3 - 6xy + y^3 + 7 for relative extrema and saddle points.

Answer 1

`(2,2)` is a minimum point of the function
`f(x,y).`

Explanation:

`f(x,y)=x^3-6xy+y^3+7`

Find the partial differentiation w.r.t.
`x` and
`y`.

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

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

`\frac {\partial^2 f}{\partial x^2}=6x`

`\frac {\partial^2 f}{\partial y^2}=6y`

`(\partial f)/(\partial xy)=-6`

Find the critical points.

`\frac {\partial f}{\partial x}=\frac {\partial f}{\partial y}=0`

`\Rightarrow 3x^2-6y=0 \;\text{and}\; -6x+3y^2=0`

`\Rightarrow 6y=3x^2\Rightarrow y=\frac{x^2}2`

`\Rightarrow -6x+\frac 34 x^4=0\Rightarrow \frac {x^3}4=2`

`\Rightarrow x^3=8\Rightarrow x=2,y=2`

Therefore,
`(2,2)` is a critical point.

Now,
`\frac {\partial^2 f}{\partial x^2}\frac {\partial^2 f}{\partial y^2}-{\frac {\partial f}{\partial xy}}^2`

`=6x6y-36=36xy-36>0 \;\text{at critical point }\; (2,2)`

Thus,
`(2,2)` is not a saddle point.

`\because (\partial^2 f)/(\partial x^2)>0 \; \text{and}\; \frac {\partial^2 f}{\partial x^2}\frac {\partial^2 f}{\partial y^2}-{\frac {\partial f}{\partial xy}}^2>0 \;\text{at point}\; (2,2)`

Hence,
`(2,2)` is a local minimum of a given function.

Answer 2

The points are "(2,2)".

Explanation:

Given value:

`f(x,y)=x^3-6xy+y^3+7`

Finding the first partial derivation which is equal to 0.

`\to f_x(x,y)=3x^2-6y...............(a)\\\\ \to f_y(x,y)=-6x+3y^2...................(b)`

solve both the above equation by equal to 0.

For equation (a)

`\to 3x^2-6y=0\\\\\to 3x^2=6y\\\\\to x^2=2y\\\\\to y=(x^2)/(2)\\\\`

For equation (b)

`\to -6x+3y^2 =0 \\\\\to 3y^2 =6x\\\\\text{put the value of y in equation b}\\\\\to 3((x^2)/(2))^2 -6x=0\\\\\to (3)/(4) x^4 -6x=0\\\\`

by solving the above value in the form of x we get:

`\to x=0\\ \to x=2\\ \to x= -1 +√(3) i\\ \to x= -1 -√(3) i\\`

by solving the above value in the form of y we get:

`\to y=0\\ \to y=2\\ \to y= -1 +√(3) i\\ \to y= -1 -√(3) i\\`

Solve the value by applying the second derivation method:

`\to f_(xx)(x,y)=6x...............(a1)\\\\ \to f_(yy)(x,y)=6y...................(b2)\\\\\to f_(xy)(x,y)=-6...................(c2)`

calculating the value of discriminate:

`d=f_(xx)(x,y) f_(yy)(x,y)-[f_(xy)(x,y)]^2`

The critical point of the given equation will be (2,2)

`d=f_(xx)(2,2) f_(yy)(2,2)-[f_(xy)(2,2)]^2\\\\`

`= (6 * 2) (6 * 2) -[(-6)^2]\\\\= (12) (12) -[36]\\\\= 144 -36\\\\= 108\\\\`