1

Find the smallest value of

$$2x^2-2xy+5y^2-6y$$

What I tried:

I used differentiation to find the answer is $-2$, and I'm looking for another way to solve this question, which is factorization. If I could make that polynomial go into the form of $a(x-t)^2+b(y-v)^2 +c$ then the answer would be $c$. Can you please help me solve it by factorization?

Community
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MCCCS
  • 1,625
  • you can not have all of $a,b,c,t,v$ are constant in the expression $a(x-t)^2+b(y-v)^2 +c$, since $2xy$ is there... – MAN-MADE Aug 23 '17 at 16:06
  • $$2 x^2-2 x y+5 y^2-6 y=2 \left(x-\frac{1}{3}\right)^2-2 \left(x-\frac{1}{3}\right) \left(y-\frac{2}{3}\right)+5 \left(y-\frac{2}{3}\right)^2-2$$ – Raffaele Aug 23 '17 at 21:47

2 Answers2

5

$$2x^2-2xy+5y^2-6y=\frac{1}{2}\left(2x-y\right)^2+\frac{1}{2}\left(3y-2\right)^2+\text{something}\,.$$

Batominovski
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2

Let $2x^2-x(2y)+5y^2-6y-k=0$

As $x$ is real, the discriminant must be $\ge0$

i.e., $(2y)^2-4\cdot2(5y^2-6y-k)\ge0$

$\iff8k\le-36y^2+48y=16-(6y-4)^2\le16$

$\iff k\le?$

We can rearrange as a quadratic equation of $y$ as well & check for the discriminant:

$$5y^2-2y(x+3)+2x^2-k=0$$

lab bhattacharjee
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