f(x,y) = 2 (x^2 + xy + 3 y + y^2)
How could that be less than 0?
How could that be less than 0?
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.