I'm solving a list of problems where I'm given an equation and I find the smallest possible value by comparing the equation to a quadratic equation and completing the square, however the next one involves another unknown $y$:
$$ x^2 – 3x + 2y^2 + 4y + 2. $$
I've been thinking about maybe somehow making $c$ equal $2y^2 + 4y + 2$?
The answer from the answerbook is: $$ x^2 – 3x + 2y^2 + 4y + 2 = \left( x – \dfrac {3}{2} \right)^2 + 2 \left( y + 1 \right)^2 – \dfrac {9}{4}. $$
Best Answer
Note that the expression can be written as $f(x)+g(y)$, where $f(x) = x^2-3x+2 $ and $g(y) = 2y(y+2)$.
The two parts are independent of each other, so we can minimize them separately.
Setting $f'(x) = 0$ gives $2x-3 = 0$, so we see that $x = \frac{3}{2}$ is the minimizer. Noting that $(x-\frac{3}{2})^2 = x^2-3x+\frac{9}{4}$, we see that we can write $f(x) = (x-\frac{3}{2})^2 - \frac{1}{4}$, and that since the square term is always non-negative, we have $f(x) \ge - \frac{1}{4}$, and $f(\frac{3}{2}) = - \frac{1}{4}$.
Similarly, setting $g'(y) = 0$ gives $y=-1$, and repeating the above, we note that $(y+1)^2 = y^2+2y+1$, so we can write $g(y) = 2(y+1)^2 -2$, and so $g(y) \ge -2$ and $g(-1) = -2$.
Putting these together, we have $f(x)+g(y) = (x-\frac{3}{2})^2 - \frac{1}{4} + 2(y+1)^2 -2 = (x-\frac{3}{2})^2 + 2(y+1)^2 - \frac{9}{4}$.
Then $f(x)+g(y) \ge - \frac{9}{4}$, and $f(\frac{3}{2}) + g(-1) = - \frac{9}{4}$.