As the title pretty much sums it up, I'm wondering what is the best way of finding the minimum value of an expression.

For instance, we have the expression $E$ defined as:$$E=x^2+4x+y^2-10y+1$$ I currently have solved it using two different methods:

  1. Forming squares of binomials
  2. Breaking it into two different expressions and finding the minimum value depending on their context - clarifications below

Solution 1:

$$E=x^2+4x+y^2-10y+1$$ $$E=(x^2+4x+4)+(y^2-10y+25) - 4 - 25 + 1$$ $$E=(x+2)^2+(y-5)^2 - 28$$

We know that $(x+2)^2 \geq 0$ and than $(y-5)^2\geq0$, and thus $E_{min} = -28$, for $x=-2$ and $y=5$.

Solution 2:

Let $E_{x} = x^2+4x$ and $E_{y}=y^2-10y$. Therefore, $E_{min}=E_{xmin}+E_{ymin}+1\tag1$

Now, for $E_{x}$ to be as small as possible, $x\leq0$, and if so: $$E_{x} = x^2-abs(4x)\implies abs(4x)>x\cdot x\tag2$$ $$(2)\implies 0\geq x\geq -4$$

Checking all four values, $E_{xmin} = -4$, for $x=-2$. The same goes for $E_{ymin}$, which results in $0\geq y\geq -10$. Checking all the values, $E_{ymin} = -25$, for $y=5$. $$(1)\implies E_{min}=-25-4+1\implies E_{min}=-28$$

(Of course for the values of x,y mentioned above)

I personally think that Solution 1 is the best, and I wonder if there is another possible solution and also which of these is recommended in this case.

  • $\begingroup$ Are you able to include the notion of gradient in your arguments? $\endgroup$ – caverac May 9 '17 at 15:21
  • $\begingroup$ Sorry for my lack of knowledge, what is a gradient @caverac ? $\endgroup$ – Mr. Xcoder May 9 '17 at 15:23
  • $\begingroup$ Well, it involves the use of partial derivatives that @Archis gave below, if you're confortable with that, then I guess you did know about gradients all along :) $\endgroup$ – caverac May 9 '17 at 15:36
  • $\begingroup$ @caverac oh, yes, I didn't realise that it was called gradient, it is different in my language, and that was the cause of my confusion $\endgroup$ – Mr. Xcoder May 9 '17 at 15:58

Hint take $E$ as $F(x,y) $ . Now take partial derivatives wrt $x $ and $y $ and set them equal to $0$ which yields two equations ie $2x+4=0,2y-10=0$ thus $x=-2,y=5$ thus minimum value of $E=-28$

  • $\begingroup$ Although I did not dive too deep into math functions, I like your solution, thank you for the new idea! $\endgroup$ – Mr. Xcoder May 9 '17 at 15:24
  • 1
    $\begingroup$ It would be quicker to solve by this method when you have one more condition imposed on $x,y $ like $ax+by+c=0$ $\endgroup$ – Archis Welankar May 9 '17 at 15:26

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