What is the minimum value of the expression $2x^2+3y^2+8x-24y+62$ for real $x$ and $y$?
I notice that this expression is a quadratic function of two variables, x and y. I also notice that the coefficients of $x^2$ and $y^2$ are both positive, which means that the function is always convex and has a unique global minimum. To find the minimum value, I need to find the values of x and y that make the partial derivatives of the function equal to zero. The partial derivative with respect to x is $4x+8$, and the partial derivative with respect to y is $6y-24$. Setting both of these equal to zero, I get $x=-2$ and $y=4$. These are the coordinates of the critical point of the function, where the minimum value occurs. To find the minimum value, I plug these values into the original expression: $2(-2)^2+3(4)^2+8(-2)-24(4)+62$. Simplifying, I get $8+48-16-96+62=6$. 