Find the critical point of the function f(x,y)=−(30x+6y^2+ln(|x+y|))
c=
To find the critical point of the function f(x, y) = -(30x + 6y^2 + ln(|x+y|)), we need to take the partial derivatives with respect to x and y and set them equal to zero.
First, let's find the partial derivative with respect to x (f_x):
f_x = -30 - 1/(x+y) (since the derivative of ln(|x+y|) with respect to x is 1/(x+y))
Next, let's find the partial derivative with respect to y (f_y):
f_y = -12y/(x+y) - 1/(x+y) (since the derivative of ln(|x+y|) with respect to y is 1/(x+y))
Setting both partial derivatives equal to zero:
-30 - 1/(x+y) = 0
-12y/(x+y) - 1/(x+y) = 0
To simplify the equations, we can multiply through by (x+y) to get rid of the fraction:
-30(x+y) - 1 = 0
-12y - (x+y) = 0
Now, we can solve these two equations simultaneously.
From the first equation, we have:
-30x - 30y - 1 = 0
-30x - 30 = 30y
x = -1/30 - y
Substituting this into the second equation:
-12y - (x+y) = 0
-12y - (-1/30 - y + y) = 0
-12y + 1/30 = 0
-12y = -1/30
y = 1/360
Now, substituting the value of y back into x = -1/30 - y:
x = -1/30 - 1/360
x = -12/360 - 1/360
x = -13/360
Therefore, the critical point of the function f(x, y) = -(30x + 6y^2 + ln(|x+y|)) is c = (-13/360, 1/360).
To find the critical point of the function f(x, y) = -(30x + 6y^2 + ln(|x+y|)), we need to take the partial derivatives with respect to both x and y and set them equal to zero.
First, let's find the partial derivative with respect to x:
∂f/∂x = -30 - (1/(x+y))*sgn(x+y)
Next, let's find the partial derivative with respect to y:
∂f/∂y = -12y - (1/(x+y))*sgn(x+y)
Setting both partial derivatives to zero:
-30 - (1/(x+y))*sgn(x+y) = 0
-12y - (1/(x+y))*sgn(x+y) = 0
Simplifying these equations, we have:
1/(x+y)*sgn(x+y) = -30
1/(x+y)*sgn(x+y) = -12y
Now, let's solve these equations simultaneously.
From the first equation, we have:
1/(x+y)*sgn(x+y) = -30
Since the sign function, sgn(x+y), can take values -1, 0, or 1, we have three cases to consider:
Case 1: sgn(x+y) = -1
In this case, -1/(x+y) = -30
=> 1/(x+y) = 1/30
=> x + y = 30
Case 2: sgn(x+y) = 0
In this case, (x+y) can take any value. We will consider this case later.
Case 3: sgn(x+y) = 1
In this case, 1/(x+y) = -30
=> x + y = -1/30
Now, let's consider case 2 where sgn(x+y) = 0.
From the second equation, we have:
1/(x+y)*sgn(x+y) = -12y
Since sgn(x+y) = 0 in this case, the equation simplifies to:
0 = -12y
=> y = 0
Now, let's substitute y = 0 into the equation x + y = -1/30:
x + 0 = -1/30
=> x = -1/30
So, the critical point of the function f(x, y) = -(30x + 6y^2 + ln(|x+y|)) is (x, y) = (-1/30, 0).