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According to a certain estimate, the depth N(t), in centimeters, of

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According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t - 5)^2 + 500 for 0 ≤ t ≤ 10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?

a) 5:30
b) 7:00
c) 7:30
d) 8:00
e) 9:00
[Reveal] Spoiler: OA

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Re: According to a certain estimate, the depth N(t), in centimeters, of [#permalink]

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New post 23 Jan 2012, 22:44
the expression is N(t)= -20(t-5)^²+500
(of course valid after 2:00 in the morning)

"the depth would be maximum" means the value of the above expression should be maximum
or the value of square term (which has a negative 20 attached to it) should be minimum i.e. zero

the square part is zero at t=5

so the time at which the depth is maximum is 2:00 + 5 hrs= 7:00 (B)

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Re: According to a certain estimate, the depth N(t), in centimeters, of [#permalink]

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New post 23 Jan 2012, 22:52
puneetkr wrote:
the expression is N(t)= -20(t-5)^²+500
(of course valid after 2:00 in the morning)

"the depth would be maximum" means the value of the above expression should be maximum
or the value of square term (which has a negative 20 attached to it) should be minimum i.e. zero

the square part is zero at t=5

so the time at which the depth is maximum is 2:00 + 5 hrs= 7:00 (B)



if -20(t-5)^²=0
then t = 5
But Why is the 500 of the equation is not considered? Show your complete calculation.
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Re: According to a certain estimate, the depth N(t), in centimeters, of [#permalink]

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New post 23 Jan 2012, 23:01
this is the complete calculation

N(t)= -20(t-5)^²+500
now for any value of t we put in; we get some negative value of -20(t-5)^² (say -x)
so our expression is now N(t)=500-x

this expression would have maximum value only when x is minimum
we know the minimum value for a square term is "zero" and (x has a square term)
and that comes when t=5
i.e. when we put t=5 here we get N(t) = 500-0 = 500

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Re: According to a certain estimate, the depth N(t), in centimeters, of [#permalink]

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Baten80 wrote:
puneetkr wrote:
the expression is N(t)= -20(t-5)^²+500
(of course valid after 2:00 in the morning)

"the depth would be maximum" means the value of the above expression should be maximum
or the value of square term (which has a negative 20 attached to it) should be minimum i.e. zero

the square part is zero at t=5

so the time at which the depth is maximum is 2:00 + 5 hrs= 7:00 (B)



if -20(t-5)^²=0
then t = 5
But Why is the 500 of the equation is not considered? Show your complete calculation.


According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t-5)^²+500 for 0≤t≤10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?

A. 5:30
B. 7:00
C. 7:30
D. 8:00
E. 9:00

Consider this: \(-20(t-5)^2\leq{0}\) hence \(500-20(t-5)^2\) reaches its maximum when \(-20(t-5)^2=0\), thus when \(t=5\). 2:00AM+5 hours=7:00AM.

Answer: B.

Hope it helps.
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Re: According to a certain estimate, the depth N(t), in centimeters, of [#permalink]

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Baten80 wrote:
According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t-5)^²+500 for 0≤t≤10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?

a) 5:30
b) 7:00
c) 7:30
d) 8:00
e) 9:00


Don't get bogged down by the dirty N(t) expression. Just think of it this way:

N(t) is a combination of two terms: a positive term (500) and a negative term (\(-20(t-5)^2\)).
To maximize N(t), I need to make the positive term as large as possible (It is a constant here so I cannot do much with it) and the absolute value of the negative term as small as possible. The smallest absolute value is 0. Can I make it 0? Yes, if I make t = 5, the negative term becomes 0 and N(t) is maximized. My answer must be 2:00 + 5 hrs i.e. 7:00.

Most of the maximum minimum questions on GMAT will require you to only think logically. The calculations involved will be minimum.
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Re: According to a certain estimate, the depth N(t), in centimeters, of [#permalink]

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New post 27 Nov 2015, 22:58
Can someone please explain why A is not the answer?

Because if I plug in t = 3.5 (which is 5.30 (3.5 after the 2:00) , it gives me the maximum value of -20(t-5)^2 expression, which will be positive. Hence, the height will be at its max of 545. Hence, shouldn't the answer be A?

Please explain the reasoning.

Thank you

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Re: According to a certain estimate, the depth N(t), in centimeters, of [#permalink]

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New post 28 Nov 2015, 08:33
herein wrote:
Can someone please explain why A is not the answer?

Because if I plug in t = 3.5 (which is 5.30 (3.5 after the 2:00) , it gives me the maximum value of -20(t-5)^2 expression, which will be positive. Hence, the height will be at its max of 545. Hence, shouldn't the answer be A?

Please explain the reasoning.

Thank you


Please read the solutions above.

Also, if t = 3.5, then 500 - 20(3.5 - 5)^2 = 455, not 545.
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Re: According to a certain estimate, the depth N(t), in centimeters, of [#permalink]

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Baten80 wrote:
According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t - 5)^2 + 500 for 0 ≤ t ≤ 10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?0


N(t)= \(-20(t - 5)^2\) + 500 ;

The red highlighted portion of the equation is the most critical part, further t = 2am + time past 2am

Now why \(-20(t - 5)^2\) is important ;

Any value of t less than 5 will result in a negative value of \(-20(t - 5)^2\)

Lets check :

Time : 3am , t = 1 { t = 2am + time past 2am }

\(-20(t - 5)^2\)

\(-20(1 - 5)^2\)

\(-20(-4)^2\)

\(-20(16)\)

\(-320\)

When put in the final equation : \(-20(t - 5)^2\) the result will be -320 + 500 => 180

Check a few more any value of t less than 5 will result in -ve value of \(-20(t - 5)^2\) and will ultimately lead to a depth of water less than 500.

Our target is getting t = 5 ; since at t = 5

\(-20(t - 5)^2\) will be 0 and the final equation \(-20(t - 5)^2\) + 500 will be maximum ; ie 500


We already know t = 2am + time past 2am

So, 7 = 2am + 5 hours.


Answer is definitely (B)
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Re: According to a certain estimate, the depth N(t), in centimeters, of [#permalink]

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Hi All,

These specific types of "limit" questions are relatively rare on Test Day, although you'll likely be tested on the concept at least once. Whenever you're asked to minimize or maximize a value, you should look to do something with the other "pieces" of the equation (usually involving maximizing or minimizing those pieces).

In the given equation, notice how you have two "parts": the -20(something) and a +500. Here, to MAXIMIZE the value of N(t), we have to minimize the "impact" that the -20(something) has on the +500. By making that first part equal 0, we'll be left with 0 + 500. Mathematically, we have to make whatever is inside the parentheses equal 0....

(T-5) = 0

T = 5

Since T represents the number of hours past 2:00am, we know that at 7:00am, the water will reach 500cm (the maximum value).

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B


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