If x and y are real numbers find the minimum value of x^2+2xy+2y^2+6y

To find the minimum value of the expression x^2 + 2xy + 2y^2 + 6y, we can use calculus. Let's take the derivative of this expression with respect to x and set it equal to 0 to find the critical points.

∂/∂x(x^2 + 2xy + 2y^2 + 6y) = 2x + 2y

Setting this equal to 0, we have:

2x + 2y = 0

Simplifying further, we get:

x + y = 0

Now let's take the derivative of the expression with respect to y and set it equal to 0.

∂/∂y(x^2 + 2xy + 2y^2 + 6y) = 2x + 4y + 6

Setting this equal to 0, we have:

2x + 4y + 6 = 0

Rearranging, we get:

2x + 4y = -6

Now we can solve this system of equations to find the values of x and y:

x + y = 0
2x + 4y = -6

Multiplying the first equation by 2, we get:

2x + 2y = 0

Now subtracting this from the second equation, we eliminate the x-term:

(2x + 4y) - (2x + 2y) = -6 - 0
2y = -6

Dividing by 2, we find:

y = -3

Plugging this back into the first equation, we find:

x + (-3) = 0
x = 3

Therefore, the critical point is (x, y) = (3, -3).

To determine if this point is a minimum or maximum, we can take the second partial derivatives. However, since this is a simple quadratic expression, we can see that the coefficient of x^2 is positive (2), which means the graph of this equation opens upwards, indicating a minimum value.

Substituting the values of x and y back into the expression, we have:

x^2 + 2xy + 2y^2 + 6y = (3)^2 + 2(3)(-3) + 2(-3)^2 + 6(-3)
= 9 - 18 + 18 - 18
= -9

Therefore, the minimum value of x^2 + 2xy + 2y^2 + 6y is -9.

To find the minimum value of the expression x^2 + 2xy + 2y^2 + 6y, we need to apply some calculus techniques.

1. Start by taking the derivative of the expression with respect to both x and y.
∂/∂x(x^2 + 2xy + 2y^2 + 6y) = 2x + 2y
∂/∂y(x^2 + 2xy + 2y^2 + 6y) = 2x + 4y + 6

2. Set each derivative equal to 0 to find the critical points.
2x + 2y = 0 (1)
2x + 4y + 6 = 0 (2)

Solve equations (1) and (2) simultaneously to find the values of x and y.

3. Subtract equation (1) from equation (2), which results in 2y + 6 = 0.
Therefore, y = -3.

4. Substitute the value of y in equation (1) to solve for x.
2x + 2(-3) = 0
2x - 6 = 0
2x = 6
x = 3.

5. The critical point is (x,y) = (3,-3).

6. Next, check whether this critical point corresponds to a minimum or maximum by finding the second derivative of the expression.

∂²/∂x²(x^2 + 2xy + 2y^2 + 6y) = 2
∂²/∂y²(x^2 + 2xy + 2y^2 + 6y) = 4

Both second derivatives are positive, indicating a minimum. Thus, the critical point (3,-3) corresponds to the minimum value of the expression.

Therefore, the minimum value of x^2 + 2xy + 2y^2 + 6y occurs when x = 3 and y = -3.

x^2+2xy+2y^2+6y

= x^2+2xy+y^2 + y^2+6y
= (x+y)^2 + y(y+6) = S

x _ y _ S

0 0 0
1 0 1
2 0 4
3 0 9
4 0 16
...
0 1 8
0 2 20
.. -------no future here, S gets bigger
try negative values

0 -6 36
1 -6 25
..
5 -6 1

5 -5 -5
1 -1 -5
2 -2 -8
3 -3 -9 <-------- LOOKS LIKE THAT IS IT
4 -4 -8

Let S = x^2 + 2xy + 2y^2 + 6y
dS/dx = 2x + 2x y' + 2y + 4y y' + 6 y'
= 0 for a min of S

y' (2x + 4y + 6) = -2x - 2y
y' = (-2x-2y)/(2x+4y+6)
could it be when y = -x ?