Find the minimum value of the expression $x^2+y^2+2x-4y+8+10x-12y$ for real $x$ and $y$.
1 answer
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.