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Re: GRE Math Challenge #54 [#permalink]
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soumya1989 wrote:
What is the maximum value of \(-3x^2 + 12x -2y^2 - 12y -39\) ?
A. -39
B. -9
C. 0
D. 9
E. 39


I don't think this is a GRE level question if this involves calculus... any algebraic way to solve...? please enlighten
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Re: GRE Math Challenge #54-value of -3x^2 + 12x -2y^2 - 12y -39 [#permalink]
Rearrange the expression to form -3*(x-2)^2-2*(y+2)^2-9.Now, the minimum value of the squares is zero, and also we should take their minimum not maximum because note that there is -3 and -2 multiplied with square expressions.So considering the square expressions zero, the final answer is -9.
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Re: GRE Math Challenge #54-value of -3x^2 + 12x -2y^2 - 12y -39 [#permalink]
i think another way to solve is by considering term containing x and y seperately and finding the amximum for that. for instance, -3x^2 +12x will be maximum when x= 2 you can try 0, value becomes 0, when you use x=1, the value becomes 9, when you try x=2 value becomes 12, now with x=3, value becomes = 9 so now going fowrward the value will decrease. so x= 2 gets maximum value.

now repeat the same procces for -2y^2 -12y which will be maximum when y= -3, i tested all negative values like -1, -2 , -3 and -4 since 12y has negative sign. the maximum value for this term will be when y=-3 and value will be 18.

now adding both values we get maximum 30 and after subtracting -39, we get maximum value of whole expression to be -9.

i hope it helps.
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Re: GRE Math Challenge #54-value of -3x^2 + 12x -2y^2 - 12y -39 [#permalink]
GreenlightTestPrep wrote:
soumya1989 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 question requires us to perform a technique called completing the square, which (in my opinion) is skirting the limits of what the GRE tests

That said here's the solution....

Take: \(-3x^2 + 12x -2y^2 - 12y -39\)

Rewrite as follows: \(-3(x^2 + 4x) - 2(y^2 + 6y) -39\)

Aside: At this point we need to recognize that \(x^2 + 4x\) will become a perfect square if we add 4, because \(x^2 + 4x + 4 = (x+2)(x+2) = (x+2)^2\)
Similarly \(y^2 + 6y + 9 = (y + 3)(y + 3) = (y + 3)^2\)

So I'm going to add the following to the algebraic expression: \(-3(x^2 + 4x + 4 - 4) - 2(y^2 + 6y + 9 - 9) -39\)
All I've done here is add and subtract 4 from the first quantity, and add and subtract 9 from the second quantity. Since the net effect is simply adding 0 the both quantities, this is a legitimate step

Now remove the \(-4\) and the \(-9\) from both brackets (they're not needed anymore)
Keep in mind that removing \(-4\) from the first set of brackets requires us to first multiply \(-4\) by \(-3\), removing \(-9\) from the second set of brackets requires us to first multiply \(-9\) by \(-2\)

When we do this we get: \(-3(x^2 + 4x + 4) + 12 - 2(y^2 + 6y + 9) + 18 -39\)

Simplify: \(-3(x^2 + 4x + 4) - 2(y^2 + 6y + 9) - 9\)

Factor: \(-3(x+2)(x+2) - 2(y+3)(y+3) - 9\)

Rewrite as follows: \(-3(x+2)^2 - 2(y+3)^2 - 9\)

At this point we need to recognize that \(-3(x+2)^2\) will be less than or equal to 0 for all values of x
So, the greatest the value of \(-3(x+2)^2\) is ZERO, and this occurs when \(x = -2\)

Likewise,\(-2(y+3)^2\) will be less than or equal to 0 for all values of y
So, the greatest the value of \(-2(y+3)^2\) is ZERO, and this occurs when \(y = -3\)

So, the greatest value of \(-3(x+2)^2 - 2(y+3)^2 - 9\) occurs when \(x = -2\) and \(y = -3\)
When we plug \(x = -2\) and \(y = -3\) into the equation, we get \(-9\)

So the greatest possible value is \(-9\)

Answer: B






"Take: \(-3x^2 + 12x -2y^2 - 12y -39\)

Rewrite as follows: \(-3(x^2 + 4x) - 2(y^2 + 6y) -39\)"

How would you get +12x if you take (-3) common.

it should be

-3(x^2 -4x) -2(y^2 +6y) -39

Please revise or explain (since you have taken the negative sign out of the parenthesis in the y part of the equation.
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GRE Math Challenge #54-value of -3x^2 + 12x -2y^2 - 12y -39 [#permalink]
Expert Reply
But if take this term it is just a semplification by 3

\(-3x^2 + 12x \)

\(-x^2\) + \(4x\)

The sign of the second term does not change regardless it is enclosed inside the parenthesis
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Re: GRE Math Challenge #54-value of -3x^2 + 12x -2y^2 - 12y -39 [#permalink]
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Carcass wrote:
But if take this term it is just a semplification by 3

\(-3x^2 + 12x \)

\(-x^2\) + \(4x\)

The sign of the second term does not change regardless it is enclosed inside the parenthesis



But he has taken -3 outside,
if within the parenthesis, -x^2 had remained, the answer would be different.

If you multiply -3(x^2 +4x)

= -3x^2 -12x

vs

3(-x^2 +4x)

=-3x^2 +12x
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Re: GRE Math Challenge #54-value of -3x^2 + 12x -2y^2 - 12y -39 [#permalink]
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Re: GRE Math Challenge #54-value of -3x^2 + 12x -2y^2 - 12y -39 [#permalink]
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