M27-12

Question

Is  xy \le \frac{1}{2} ?

(1)  x^2+y^2=1 
 
(2)  x^2-y^2=0

Bunuel · Accepted Answer

Official Solution:

Is  xy \le \frac{1}{2} ? 
 
(1)  x^2+y^2=1 . 
 
Recall that  (x-y)^2 \ge 0  (since the square of any number is always non-negative). If we expand this, we get  x^2-2xy+y^2 \ge 0 . Given that  x^2+y^2=1 , we can substitute this into our equation, resulting in  1-2xy \ge 0 . From this, we can deduce that  xy \le \frac{1}{2} . Sufficient.
 
(2)  x^2-y^2=0 . 
 
If we rearrange this and take the square root of both sides, we get  |x|=|y| . However, this does not provide us with enough information to answer whether  xy \le \frac{1}{2} . Not sufficient.

Answer: A

eureka750 · Answer

The same applicable to (x+y)^2 ==> x^2 + y^2 + 2 x*y is greter than or equal to ZERO. 
1 + 2 x*y is greter than or equal to ZERO.
xy greter than or equal -1/2...This can prove A is insufficient, right ???

Bunuel · Answer

You need to consider  (x-y)^2\geq{0}  to get sufficiency.

BCCLSN · Answer

Why did you start from : (x-y)^2>=0 ??

harshalnamdeo88 · Answer

Please explain why do we need to consider only (x-y)^2 to get sufficiency, as also in (x+y)^2 we use x^2 + y^ 2=1 premise, hence its not in all cases that(A) results in single conclusion or sufficiency.

shrutikasat · Answer

yeah even i have a same doubt why not (x+y)^2 as harshalnamdeo88...

Bunuel · Answer

The key is that if you look at  (x+y)^2 \ge 0 , you'll find  xy\geq{-\frac{1}{2} } . But this isn't really helpful. It's true that both  xy\geq{-\frac{1}{2} }  (from  (x+y)^2 \ge 0 ) and  xy \le \frac{1}{2}  (from  (x-y)^2 \ge 0 ) are correct. So, this means  -\frac{1}{2} \leq{xy} \leq{\frac{1}{2} } . But only the method shown in the original solution gets us the answer we want. The other way just gives us an inequality that doesn't help much.

Hope it's clear.

spence11 · Answer

This is a hard question, in particular because knowing to use the square of a difference is a tricky detail.

You can also use the fact that the maximum product of  ab  given that  a+b=c  is the set of numbers where  a  and  b  are nearest each other and still sum to  c .

For example, if  a+b=20 , the maximum product of  ab=10*10=100

In this case the maximum product of  ab  where  a+b=1  is when  a=b=\frac{1}{2}

In this case  x^2=a  and  y^2=b , so the the maximum product of  xy  is  \sqrt{\frac{1}{2}}*\sqrt{\frac{1}{2}}=\frac{1}{2}

Answer A

AndrewN · Answer

Hello, everyone. If it helps, I thought I would add a graphical solution. Notice that the equation in Statement (1) conforms to that for a circle: (x-h)^2+(y-k)^2=r^2 in which (h, k) is the center of the circle and r is the radius. This circle would fit the equation thus: (x - 0)^2 + (y-0)^2=1^2 x^2 + y^2=1 This is the very unit circle that many trigonometry and pre-calculus students (at least, in the US) are taught, and recognizing it here can save a lot of time and effort. Screen Shot 2022-04-18 at 14.26.33.png Where it touches the circle, the green segment represents the maximum value for xy , since, in quadrants I and III, the two unknowns will have the same sign. The maximum value for the product will be found at 45 (or 225) degrees, equidistant from the x- and y- axes. (You can use a 45-45-90 special right triangle if you like. You know that in the corresponding side ratios x , x , and x √2, respectively, 1, the radius of the circle, is equivalent to x √2. This leads to a leg length of 1/√2 or, when the denominator is rationalized, √2/2.) Thus, the maximum value of xy is \frac{1}{\sqrt{2}}*\frac{1}{\sqrt{2}} = \frac{1}{2} Since 1/2 is equal to 1/2, but we know the product cannot be any greater, Statement (1) is SUFFICIENT . Statement (2) should take little more than a glance to see that the equation will prove insufficient on its own. Just as Bunuel has done above, start by rearranging. x^2-y^2=0 x^2 = y^2 |x| = |y| On its own, the statement tells us nothing about the nature of xy , relative to 1/2. The answer must be (A). Although knowledge of the circle equation and the unit circle is not required for the GMAT™, it can come in handy in unexpected ways, sometimes on tougher questions such as this one. Good luck with your studies. - Andrew

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

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