If X and Y are real numbers Determine the minimum value of 2x^2+2xy+2y^2+6y how do i separate the two to get two simultaneous equations?do i use the inequality sign > ?
f(x,y) = 2 (x^2 + xy + 3 y + y^2)
How could that be less than 0?
It was given to me like that I also wanted to use the >sign. Please help me because I'm not sure of how to separate it
To determine the minimum value of the expression 2x^2 + 2xy + 2y^2 + 6y, you do not need to separate it into two equations. Instead, you can use calculus to find the minimum point.
To find the minimum value, you need to take the derivative of the expression with respect to both x and y, and then find the critical points where both derivatives are equal to zero.
Let's start by taking the derivative with respect to x:
d/dx (2x^2 + 2xy + 2y^2 + 6y) = 4x + 2y
Next, take the derivative with respect to y:
d/dy (2x^2 + 2xy + 2y^2 + 6y) = 2x + 4y + 6
To find the critical points, set both derivatives equal to zero and solve the system of equations:
4x + 2y = 0 (Equation 1)
2x + 4y + 6 = 0 (Equation 2)
Now, you have two simultaneous equations that you can solve to find the values of x and y at the critical points.
To solve this system of equations, you can use various methods such as substitution or elimination.
The inequality sign ">" is not needed in this case, as you are looking for the minimum value, not the inequality relationship between two expressions.