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 ?