For f(x,y)=x^2+y^2-xy+6y, determine the values for x and y such that fx(x,y)=0 and fy(x,y)=0 simultaneously.
To determine the values for x and y such that fx(x, y) = 0 and fy(x, y) = 0 simultaneously, we need to find the partial derivatives of the function f(x, y) with respect to x and y, and set them equal to zero.
Let's start by finding the partial derivative of f(x, y) with respect to x, which is denoted as fx(x, y):
To find fx(x, y), we differentiate f(x, y) with respect to x, treating y as a constant.
Differentiating x^2 with respect to x gives us 2x.
Differentiating -xy with respect to x gives us -y.
Differentiating 6y with respect to x gives us 0 (since y does not contain x).
So, fx(x, y) = 2x - y.
Next, let's find the partial derivative of f(x, y) with respect to y, denoted as fy(x, y):
To find fy(x, y), we differentiate f(x, y) with respect to y, treating x as a constant.
Differentiating x^2 with respect to y gives us 0 (since x does not contain y).
Differentiating -xy with respect to y gives us -x.
Differentiating 6y with respect to y gives us 6.
So, fy(x, y) = -x + 6.
Now, to solve the system of equations fx(x, y) = 0 and fy(x, y) = 0 simultaneously, we set the partial derivatives equal to zero:
2x - y = 0 (Equation 1)
-x + 6 = 0 (Equation 2)
From Equation 1, we can solve for y in terms of x:
y = 2x (Equation 3)
Substituting Equation 3 into Equation 2, we have:
-x + 6 = 0
Solving for x, we find:
x = 6
Now, substituting the value of x into Equation 3, we can find the corresponding value of y:
y = 2(6) = 12
Therefore, the values for x and y that satisfy the equations fx(x, y) = 0 and fy(x, y) = 0 simultaneously are x = 6 and y = 12.