Problem Solving

What is the maximum value of −3x^2+12x−2y^2−12y−39?

A. −39
B. −9
C. 0
D. 9
E. 39

OA:B

Mo2men wrote:What is the maximum value of −3x^2+12x−2y^2−12y−39?

A. −39
B. −9
C. 0
D. 9
E. 39

This type of question is typically solved by COMPLETING THE SQUARE -- a process that is beyond the scope of the GMAT.
Feel free to ignore this problem.

A GMAT-friendly approach:

Maximize the value of −3x² + 12x:
If x=0, then −3x^2 + 12x = 0.
If x=1, then −3x^2 + 12x = 9.
If x=2, then −3x^2 + 12x = 12.
If x=3, then −3x^2 + 12x = 9.
The results above indicate that the greatest possible value of −3x²+12x is 12.

Maximize the value of −2y² − 12y = -2(y² + 12y).
Here, the value will be maximized if the expression in red is NEGATIVE.
Test negative values in −2y²−12y.
If y=-1, then −2y² − 12y = 10.
If y=-2, then −2y² − 12y = 16.
If y=-3, then −2y² − 12y = 18.
If y=-4, then −2y² − 12y = 16.
The results above indicate that the greatest possible value of −2y²−12y is 18.

Thus:
The greatest possible value of −3x²+12x−2y²−12y−39 = (greatest possible value of −3x²+12x) + (greatest possible value of −2y²−12y) − 39 = 12 + 18 - 39 = -9.

The correct answer is B.

Brent@GMATPrepNow wrote:
−3x² + 12x − 2y² − 12y − 39 = −3(x² - 4x) − 2(y² − 6y) − 39

Check italicized for sign consistency

-3x² + 12x - 2y² - 12y - 39 =>

-1 * (3x² - 12x + 2y² + 12y + 39) =>

-1 * (3*(x² - 4x) + 2*(y² + 6y) + 39) =>

-1 * (3*((x - 2)² - 4) + 2((y + 3)² - 9) + 39) =>

-1 * (3(x - 2)² - 12 + 2(y + 3)² - 18 + 39) =>

-1 * (3(x - 2)² + 2(y + 3)² + 9) =>

-9 - 3(x - 2)² - 2(y + 3)²

You'll maximize that if you aren't subtracting ANYTHING from -9, i.e. if x = 2 and y = -3. So the max is -9 when x = 2 and y = -3.