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What is the maximum value of the expression 22/(4x^4 - 32x^2 + 75)

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What is the maximum value of the expression 22/(4x^4 - 32x^2 + 75)  [#permalink]

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New post 30 Dec 2016, 09:47
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What is the maximum value of the expression \(\frac{22}{4x^4 - 32x^2 + 75}\)

A) 1
B) 10/9
C) 2
D) 23/11
E) 25/11
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Re: What is the maximum value of the expression 22/(4x^4 - 32x^2 + 75)  [#permalink]

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New post 30 Dec 2016, 10:33
Rocky1304 wrote:
what is the maximum value of the expression
\(22/(4x^4-32x^2+75)\)

A) 1
B) 10/9
C)2
D) 23/11
E) 25/11


\(\frac{22}{(4x^4-32x^2+75)}\)

This expression will take its maximum value when \(4x^4-32x^2+75\) will take its minimum value.

\(4x^4-32x^2+75 = (2x^2)^2 + 2*2x^2*8 + 64 + 11 = (2x^2 - 8)^2 + 11\)

If we get negative or positive value of \(2x^2 - 8\) it will be squared and become positive, hence increasing our denominator and decreasing value of our fraction. The only possible minimum can be achieved when \(2x^2 - 8 = 0\)

\(2x^2 - 8 =0\)

\(x^2 = 4\)

\(x= +/- 2\)

Answer C
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Re: What is the maximum value of the expression 22/(4x^4 - 32x^2 + 75)  [#permalink]

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New post 30 Dec 2016, 23:02
vitaliyGMAT wrote:
Rocky1304 wrote:
what is the maximum value of the expression
\(22/(4x^4-32x^2+75)\)

A) 1
B) 10/9
C)2
D) 23/11
E) 25/11


\(\frac{22}{(4x^4-32x^2+75)}\)

This expression will take its maximum value when \(4x^4-32x^2+75\) will take its minimum value.

\(4x^4-32x^2+75 = (2x^2)^2 + 2*2x^2*8 + 64 + 11 = (2x^2 - 8)^2 + 11\)

If we get negative or positive value of \(2x^2 - 8\) it will be squared and become positive, hence increasing our denominator and decreasing value of our fraction. The only possible minimum can be achieved when \(2x^2 - 8 = 0\)

\(2x^2 - 8 =0\)

\(x^2 = 4\)

\(x= +/- 2\)

Answer C



Hey,

I don't think there was any need to find the value of x in this case. As correctly pointed by you the value of the expression will be maximum when the denominator is minimum and to do that we need to need ensure that \(2x^2 - 8 = 0\)

Thus the maximum value of the expression is :

\(22/(4x^4-32x^2+75)\)
\(={22}{(2x^2 - 8)^2 + 11}\)
\(=22/11\)
\(= 2\)

I think it just a coincidence that the value of x and the value of the expression is 2. And maybe that is why you did not notice that my mistake you found the value of x instead of the value of the expression. Just thought to let you know. :)


Thanks,
Saquib
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Senior Manager
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Joined: 13 Oct 2016
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Re: What is the maximum value of the expression 22/(4x^4 - 32x^2 + 75)  [#permalink]

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New post 31 Dec 2016, 00:39
EgmatQuantExpert wrote:
vitaliyGMAT wrote:
Rocky1304 wrote:
what is the maximum value of the expression
\(22/(4x^4-32x^2+75)\)

A) 1
B) 10/9
C)2
D) 23/11
E) 25/11


\(\frac{22}{(4x^4-32x^2+75)}\)

This expression will take its maximum value when \(4x^4-32x^2+75\) will take its minimum value.

\(4x^4-32x^2+75 = (2x^2)^2 + 2*2x^2*8 + 64 + 11 = (2x^2 - 8)^2 + 11\)

If we get negative or positive value of \(2x^2 - 8\) it will be squared and become positive, hence increasing our denominator and decreasing value of our fraction. The only possible minimum can be achieved when \(2x^2 - 8 = 0\)

\(2x^2 - 8 =0\)

\(x^2 = 4\)

\(x= +/- 2\)

Answer C



Hey,

I don't think there was any need to find the value of x in this case. As correctly pointed by you the value of the expression will be maximum when the denominator is minimum and to do that we need to need ensure that \(2x^2 - 8 = 0\)

Thus the maximum value of the expression is :

\(22/(4x^4-32x^2+75)\)
\(={22}{(2x^2 - 8)^2 + 11}\)
\(=22/11\)
\(= 2\)

I think it just a coincidence that the value of x and the value of the expression is 2. And maybe that is why you did not notice that my mistake you found the value of x instead of the value of the expression. Just thought to let you know. :)


Thanks,
Saquib
Quant Expert
e-GMAT

To practise ten 700+ Level Number Properties Questions attempt the The E-GMAT Number Properties Knockout



Image


Hi Saquib
Thanks for your kind advice. It was way beyond midnight and I did not put the solution to its logical end. What I’ve found in fact is the value of x that will maximize our fraction. In general the proper way to solve questions like that, as it’s done in analysis, is to find first and second derivatives of denominator and extreme values of our variable, put them into fraction to find its max or min value. But, calculus is beyond gmat, that’s why we have this case where we can simplify our expression to “perfect square plus 11” and, as you’ve correctly mentioned, it’s quite enough to take our perfect square equal to zero. I just had to put that x into our fraction, but that was unnecessary, 22/11 will be our answer. In general, this won’t do and we’ll need to go the long way.
Thanks again!
Merry holidays
Vitaliy
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Re: What is the maximum value of the expression 22/(4x^4 - 32x^2 + 75)  [#permalink]

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New post 19 Sep 2018, 11:14
In order to maximize the expression the denominator should be minimum.

For a quadratic equation to be minimum the d/dx[(4x^4-32x^2+75)]=16x^3 - 64x=0
Which gives x=+2,-2
Substitute this in the original expression and we get answer as (c).
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Re: What is the maximum value of the expression 22/(4x^4 - 32x^2 + 75)  [#permalink]

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New post 19 Sep 2018, 16:31
1
Can someone please break down how

\(4x^4\)−\(32x^2\)+75 becomes (\(2x^2\)−8)\(^{2}\)+11 ?

Thanks!
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Re: What is the maximum value of the expression 22/(4x^4 - 32x^2 + 75)  [#permalink]

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New post 20 Sep 2018, 02:14
2
RichardSaunders wrote:
Can someone please break down how

\(4x^4\)−\(32x^2\)+75 becomes (\(2x^2\)−8)\(^{2}\)+11 ?

Thanks!


you need to think about \((a-b)^2 = a^2 - 2ab + b^2\)
\(4x^4−32x^2+75\)

\(4x^4\) can be weritten as \(( 2 x^2 ) ^ 2\) = \(a^2\)

\(- 32x^2\) = 2 * \(2x^2\) * (-8) = 2ab (you need 32/4 = 8)

\(b^2 = -8^2 = 64\) (you can separate 75 into 64 + 11)

Therefore \((2x^2 - 8)^ + 11\)
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Re: What is the maximum value of the expression 22/(4x^4 - 32x^2 + 75)  [#permalink]

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New post 20 Sep 2018, 04:33
Thanks pandeyashwin, very simple/clear
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Re: What is the maximum value of the expression 22/(4x^4 - 32x^2 + 75) &nbs [#permalink] 20 Sep 2018, 04:33
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