daniel suriya

daniel suriya

Answered question

2022-07-19

Answer & Explanation

Nick Camelot

Nick Camelot

Skilled2023-06-05Added 164 answers

To find the critical points of the function f(x,y)=x+2x2+2xy2+y4, we need to determine the values of x and y where the partial derivatives of f with respect to x and y are both zero.
Let's start by finding the partial derivative of f with respect to x:
fx=1+4x+2y2
Now, let's find the partial derivative of f with respect to y:
fy=4xy+4y3
To find the critical points, we set both partial derivatives equal to zero and solve the resulting system of equations:
1+4x+2y2=0 ...(1)
4xy+4y3=0 ...(2)
From equation (2), we can factor out 4y:
4y(x+y2)=0
This equation is satisfied if either 4y=0 or x+y2=0.
If 4y=0, then y=0.
If x+y2=0, then x=y2.
Substituting y=0 into equation (1), we find:
1+4x+2(0)2=0
1+4x=0
x=14
Therefore, we have found one critical point at (x,y)=(14,0).
Now, let's consider the case when y0. From equation (2), we can divide both sides by 4y:
x+y2=0
Solving this equation for x, we find:
x=y2
Therefore, for any y0, we have a critical point at (x,y)=(y2,y).
To classify the critical points, we need to determine the nature of these points. To do this, we can use the second partial derivative test. We calculate the second partial derivatives and evaluate them at each critical point.
The second partial derivatives of f are:
2fx2=4
2fy2=12y2
2fxy=4y
Evaluating the second partial derivatives at each critical point, we have:
At (x,y)=(14,0):
2fx2=4
2fy2=0
2fxy=0
The discriminant of the Hessian matrix is:
D=(2fx2)(2fy2)(2fxy)2=(4)(0)(0)2=0
Since the discriminant is zero, the second derivative test is inconclusive.
At (x,y)=(y2,y):
2fx2=4
2fy2=12y2
2fxy=4y
The discriminant of the Hessian matrix is:
D=(2fx2)(2fy2)(2fxy)2=(4)(12y2)(4y)2=48y216y2=32y2
The discriminant depends on the value of y. If y=0, the discriminant is zero, indicating an inconclusive test. For y0, the discriminant is positive, indicating a local minimum or maximum.
To summarize, the critical points of the function f(x,y)=x+2x2+2xy2+y4 are:
1. (x,y)=(14,0) - Inconclusive test
2. (x,y)=(y2,y) for y0 - Local minimum or maximum

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