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Maximum value of
[#permalink]
28 Sep 2021, 00:31
28 Sep 2021, 00:31
Expert Reply
00:00
Question Stats:
88% (00:35) correct
11% (00:27) wrong based on 9 sessions
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Quantity A |
Quantity B |
Maximum value of \(\frac{1}{4+y^2}\) |
\(\frac{1}{4 }\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Source: manhattanreview
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Maximum value of
[#permalink]
28 Sep 2021, 09:14
28 Sep 2021, 09:14
1
Carcass wrote:
Quantity A |
Quantity B |
Maximum value of \(\frac{1}{4+y^2}\) |
\(\frac{1}{4 }\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Source: manhattanreview
In order to maximize the value of \(\frac{1}{4+y^2}\), we must minimize the value of the denominator, \(4 + y^2\)
Since \(y^2\) is greater than or equal to 0 for all values of \(y\), we can see that \(4 + y^2\) is minimized when \(y = 0\)
When \(y = 0\), \(\frac{1}{4+y^2}=\frac{1}{4+0^2}=\frac{1}{4+0}=\frac{1}{4}\)
So, we have:
QUANTITY A: \(\frac{1}{4 }\)
QUANTITY B: \(\frac{1}{4 }\)
Answer: C
gmatclubot
Re: Maximum value of [#permalink]
28 Sep 2021, 09:14
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