Question

Solve the function f(x,y) = x^3 + 3y^3 + 3x^2 + 3y^2 +24
At what point Hessian matrix is (a) positive definite. (b) negative definite (c) indefinite ? Extreme points of given function is ? Write the value of function at (-2,-2/3)?

Answer 1

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Answer 2

Concept:

A derivative is a contract between two or more parties in which the contract's value is decided by an agreed-upon underlying financial asset (such as security) or combination of assets (like an index).

Given:

$f(x,y)=x^{3}+3y^{3}+3x^{2} +3y^{2} +24$

To Find:

At what point Hessian matrix is (a) positive definite. (b) negative definite (c) indefinite? Extreme points of the given function are? Write the value of the function at (-2,-2/3)

Solution:

A derivative is a contract between two or more parties in which the contract's value is decided by an agreed-upon underlying financial asset (such as security) or combination of assets (like an index).

Let $f(x,y)=x^{3}+3y^{3}+3x^{2} +3y^{2} +24$

$fx=3x^{2} +3y^{2}-6x-0+0\\fxx=6x-6\\Also, fy=0+6xy-0-6y+0\\fyy=6x-6\\fxy=0+6y-0$

Put,

$fx=0 and fy=0\\3x^{2} +3y^{2}-6x=0\\ x^{2} +y^{2}-2x=0\\ And 6xy-6y=0\\6y(x-1)=0\\y=0 and x=1$

Case:1 when x=1

$1^{2}+y^{2} -2(1)=0\\ y^{2}-1=0\\ y=1, -1$

When y=0

$x^{2} +0-2x=0\\x(x-2)=0\\x=0, x=2$

∴The stationery points are (1,1), (1,-1), (0,0), and (2,0)

Now, at (-2 and -2/3)

So,

$6y(x-1)\\6*(\frac{-2}{3}) (-2-1)\\ -\frac{12}{3} (-3)\\-4*(-3)\\12$

∴The value of the function at (-2,-2/3) is 12.

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