happyface101 wrote:
iwantstanford wrote:
How did you know to split the -39 into 3 parts: -12, -18, and -9?
Hi Experts - can you please clarify how to recognize that we need to split -39 to solve?
I'm not sure how to recognize that this is the approach. Also, how do you know what to split -39 into?
The trick with these questions that ask you about max/min of a second degree polynomial is that you will invariably get a value of the form (a\(\pm\)b) wherein, the minimum value of any square = 0. Thus try to create perfect squares out of the given polynomials.
Realize that \((a\pm b)^2 = a^2 \pm 2*a*b + b^2\).
Given expression: \(−3x^2+12x−2y^2−12y−39\) ---> try to create perfect squares :
\((-3x^2+12x) + (-2y^2-12y)-39 ---> -3 (x^2-4x) -2 (y^2+6y) -39 ----> 3 (x^2-4x+4-4) -2 (y^2+6y+9-9) -39\)
---> \(3 (x^2-4x+4-4) -2 (y^2+6y+9-9) -39\) ---> \(-3[(x-2)^2-4] -2 [(y+3)^2-9]-39\) ----> \(-3(x-2)^2+12 -2(y+3)^2+18-39\)
----> \(-3(x-2)^2+12 -2(y+3)^2+18-39\) --->\(-3(x-2)^2-2(y+3)^2-9\)
As mentioned above, the minimum value of a perfect square = 0 --> to maximize the given expression,
you need to put \((x-2)^2 =0\) and\((y+3)^2=0\) ---> giving you 0+0-9=-9 as the maximum value.
Any other values of\((x-3)^2\) and \((y+3)^2\)will give you a smaller negative value hence 'minimizing' instead of maximizing the expression.
Hope this helps.