Find the minimum value of the expression $x^2+y^2+2x-4y+8+10x-12y$ for real $x$ and $y$.
To minimize the expression $x^2+y^2+2x-4y+8+10x-12y$, we complete the square in $x$ and $y$ separately in order to write the expression in the form $(x-a)^2+(y-b)^2+c$. The square that completes the square in $x$ is $(x+1)^2-1=(x+1)^2+0^2-1$ and the square that completes the square in $y$ is $(y-2)^2-4=(y-2)^2+0^2-4$, so the given expression is equal to $$(x+1)^2+0^2-1+(y-2)^2+0^2-4+10x-12y$$$$=(x+1)^2+(y-2)^2+10x-12y-5.$$ To minimize the expression, we minimize each of the terms $(x+1)^2$, $(y-2)^2$, $10x$, and $-12y$ separately. By the Trivial Inequality, $(x+1)^2\geq0$ for all $x$. Since the minimum value of $(x+1)^2$ is $0$ (which occurs at $x=-1$), the minimum value of $(x+1)^2$ is $0$. By the Trivial Inequality, $(y-2)^2\geq0$ for all $y$. Since the minimum value of $(y-2)^2$ is $0$ (which occurs at $y=2$), the minimum value of $(y-2)^2$ is $0$. The term $10x$ takes on its minimum value when $x$ is a global minimum, which does not occur for $x$. Therefore, the minimum value of $10x$ is $-\infty$. The term $-12y$ takes on its maximum value when $y$ is a global maximum, which does not occur for $y$. Therefore, the maximum value of $-12y$ is $\infty$. The sum of any specific finite number and $-\infty$ is $-\infty$, and the sum of any specific finite number and $\infty$ is $\infty$. therefore the minimum of $10x$ and $-12y$The minimum value of $x^2+y^2+2x-4y+8+10x-12y$ is $1+0-5=\boxed{-4}$. From our analysis, we see that $(x,y)=(-1,2)$ results in equality, so our answer is reasonable.
To find the minimum value of the expression $x^2+y^2+2x-4y+8+10x-12y$, we can complete the square for both $x$ and $y$ terms.
For the $x$ terms, we complete the square by adding $(2/2)^2 = 1$ inside the parentheses:
\[x^2 + 2x + 10x = (x^2 + 2x + 1) + 10x - 1 = (x+1)^2 + 10x - 1.\]
For the $y$ terms, we complete the square by adding $(-4/2)^2 = 4$ inside the parentheses:
\[y^2 - 4y - 12y = (y^2 - 4y + 4) - 4^2 = (y - 2)^2 - 16.\]
Now, our expression becomes:
\[(x+1)^2 + 10x - 1 + (y-2)^2 - 16 + 8.\]
Simplify this expression as:
\[(x+1)^2 + (y-2)^2 + 10x -24.\]
To minimize the expression, we want to minimize the sum of squares $(x+1)^2$ and $(y-2)^2$, while minimizing $10x$.
The minimum value of $(x+1)^2$ occurs when $x = -1$, and its value is 0. Similarly, the minimum value of $(y-2)^2$ is 0 and occurs when $y=2$.
Thus, the minimum value of the expression is when $x = -1$ and $y = 2$.
Substituting these values, we get:
\[(x+1)^2 + (y-2)^2 + 10x -24 = 0 + 0 + 10(-1) - 24 = -10 - 24 = \boxed{-34}.\]
To find the minimum value of the expression $x^2+y^2+2x-4y+8+10x-12y$, we can use the concept of completing the square.
First, let's group the terms that involve $x$ and $y$ separately:
$x^2+y^2+2x-4y+8+10x-12y$
$= (x^2+2x+10x)+(y^2-4y-12y)+8$
Next, we'll complete the square for the $x$ terms. To complete the square for $x$, we need to add $\left(\frac{2}{2}\right)^2=1$ inside the parentheses:
$= (x^2+2x+1+10x)+(y^2-4y-12y)+8-1$
$= (x+1)^2+10x+(y^2-4y-12y)+8-1$
Similarly, we'll complete the square for the $y$ terms. To complete the square for $y$, we need to add $\left(\frac{-4}{2}\right)^2=4$ inside the parentheses:
$= (x+1)^2+10x+(y^2-4y+4-4-12y)+8-1$
$= (x+1)^2+10x+(y-2)^2-4(y+1)+8-1$
$= (x+1)^2+10x+(y-2)^2-4y-4+8-1$
$= (x+1)^2+10x+(y-2)^2-4y+3$
Now, let's rewrite the expression:
$= (x+1)^2+10x+(y-2)^2-4y+3$
We can see that this expression can be written as:
$(x+1)^2+10x+(y-2)^2-4y+3 = (x+1)^2 + (y-2)^2 + 10x - 4y + 3$
We can now compare this expression to the standard form of a quadratic function, which is $ax^2+bx+c$. We can see that the expression has a coefficient of 1 for both $(x+1)^2$ and $(y-2)^2$. Therefore, we can rewrite the expression as:
$(x+1)^2 + (y-2)^2 + 10x - 4y + 3 = (x+1)^2 + (y-2)^2 + 10x - 4y + 3(1)$
Now, we can see that the constant term, $c$, in the standard form is 3(1). Since $3(1)$ is positive, the minimum value of the expression occurs when $(x+1)^2$ and $(y-2)^2$ are both equal to 0. This means $x+1=0$ and $y-2=0$.
Solving these equations, we find that $x=-1$ and $y=2$.
Therefore, the minimum value of the expression is $0$ when $x=-1$ and $y=2$.