How do you find the smallest possible value of aan equation with two unknowns?

Question

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: $$ x^2 - 3x + 2y^2 + 4y + 2 = \left( x - \dfrac {3}{2} \right)^2 + 2 \left( y + 1 \right)^2 - \dfrac {9}{4}. $$

Answer 1

EDIT: $$ x^2 - 3x + 2y^2 + 4y + 2$$ can be re-written as $$ x^2-2*x\frac{3}{2}+\frac{9}{4}+2(y^2+2y+1)-\frac{9}{4}$$ Obeserve that the first two terms in the expression are of the form $$a^2+2a.b+b^2=(a+b)^2$$ Hence we get $$ x^2 - 3x + 2y^2 + 4y + 2=x^2-2x\frac{3}{2}+\frac{9}{4}+2(y^2+2y+1)-\frac{9}{4}$$ $$\Rightarrow x^2 - 3x + 2y^2 + 4y + 2=(x-\frac{3}{2})^2+2(y+1)^2-\frac{9}{4}$$

On the RHS we have the sum of two squares is is . The minimum value of each square term will be 0 as a square term cannot be negative in $IR$. Hence the minimum value of the given expression is -$\frac{9}{4}$ when the two square terms are 0(at x=$\frac{3}{2}$ and y=-1).

Answer 2

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 $f(\frac{3}{2}) + g(-1) = - \frac{9}{4}$.