Minimum value of an expression

Rahul@gurome wrote:

thp510 wrote:...
Not getting it. What does the equation look like before you get to 4x=4 or 6y=12 using the differentiation rule. I'm still stuck at the following:

2(x^2-2x)+3(y^2-4y)+18

To use the differentiation rule, you need to know differential calculus which is not a part of GMAT. You can proceed as I mentioned earlier. Let's do it again. To minimize the expression, we have to minimize each term of it. Simple algebraical method to do it is to rearrange the terms in such a way that we get square terms. This is because minimum value of a square term is zero.

# 2x² + 3y² - 4x - 12y + 18
= 2(x² - 2x) + 3(y² - 4y) + 18 .................... What you've done
= 2(x² - 2x + 1) + 3(y² - 4y + 4) + 4
= 2(x - 1)² + 3(y - 2)² + 4

For minimum value of the expression, 2(x - 1)² and 3(y - 2)² must be minimum, i.e. equal to zero. Thus minimum value of the expression is 4.

christitk wrote:Can you please explain how the 18 became 4 in that equation??

Rahul@gurome wrote:

thp510 wrote:...
Not getting it. What does the equation look like before you get to 4x=4 or 6y=12 using the differentiation rule. I'm still stuck at the following:

2(x^2-2x)+3(y^2-4y)+18

To use the differentiation rule, you need to know differential calculus which is not a part of GMAT. You can proceed as I mentioned earlier. Let's do it again. To minimize the expression, we have to minimize each term of it. Simple algebraical method to do it is to rearrange the terms in such a way that we get square terms. This is because minimum value of a square term is zero.

# 2x² + 3y² - 4x - 12y + 18
= 2(x² - 2x) + 3(y² - 4y) + 18 .................... What you've done
= 2(x² - 2x + 1) + 3(y² - 4y + 4) + 4
= 2(x - 1)² + 3(y - 2)² + 4

For minimum value of the expression, 2(x - 1)² and 3(y - 2)² must be minimum, i.e. equal to zero. Thus minimum value of the expression is 4.