Bunuel wrote:
4. What is the maximum value of -3x^2 + 12x -2y^2 - 12y - 39 ?
A. -39
B. -9
C. 0
D. 9
E. 39
\(-3x^2 + 12x -2y^2 - 12y - 39=-3x^2 + 12x-12-2y^2 - 12y-18-9=-3(x^2-4x+4)-2(y^2+6y+9)-9=\)
\(=-3(x-2)^2-2(y+3)^2-9\).
So, we need to maximize the value of \(-3(x-2)^2-2(y+3)^2-9\).
Since, the maximum value of \(-3(x-2)^2\) and \(-2(y+3)^2\) is zero, then the maximum value of the whole expression is \(0+0-9=-9\).
Answer: B.
Hi Bunuel,
Just as a matter of fact can you please explain that how you reached to factorization. In other words how a candidate will decide how to split -39 between (X,Y).
Your reply is appreciated !!
Rgds,
TGC !!
I completed the squares for -3x^2 + 12x - ... and for -2y^2 - 12y-... So, I asked myself what do I need there in order to have (a+b)^2.