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).
Find the critical point of the function f(x,y)=−(30x+6y^2+ln(|x+y|))
c=
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