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

Question

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

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

KarishmaB · Accepted Answer

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.

Bunuel · Answer

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

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.

puneetkr · Answer

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)

MHIKER · Answer

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

MHIKER · Answer

A similar problem:  https://gmatclub.com/forum/an-object-thrown-directly-upward-88791.html

EMPOWERgmatRichC · Answer

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).

Final Answer:  B

GMAT assassins aren't born, they're made, 
Rich

ScottTargetTestPrep · Answer

Since -20(t - 5)^2, will be a nonpositive number, its maximum value is 0 when t = 5, and the maximum value of the function will then be:

N(5) = -20(5 - 5)^2 + 500 = 500

Thus, the maximum depth is at 2am + 5 hours = 7am.

Answer: B

rjacobson99 · Answer

Can someone explain why / how you would know not to simplify the original provided equation (-20(t-5)^2 + 500)? When you simplify the equation, you seems to get a different answer. Please see work below:

-20(t-5)^2 + 500
-20(t^2 - 10t + 25) + 500 <-- foil
-20t^2 + 200t - 500 + 500 <-- distributed the -20
-20t^2 + 200t
20t(t + 10t) <-- simplified formula

Given the 20t(t+10t) is simplified correctly, then the water level will continue to grow with every hour with the latest hour being the maximum. This is different than the answer provided, where t is at the max once 5 hours past.

EMPOWERgmatRichC · Answer

Hi rjacobson99,

In the last "step" of you work, you have not properly accounted for the 'minus' sign. If you want to factor out "20t", then that's fine, but here's what you would be left with:

-20t^2 + 200t
(20t)(-t + 10)
(20t)(10 - t)

The maximum result will occur when t = 5.

GMAT assassins aren't born, they're made,
Rich

rjacobson99 · Answer

Hi Rich,

In the correct version of the formula (20t)(10-t), is there a way to determine the inflection point of t = 5 without having to plug in values from t = 1 to 6? I.e. is there a shortcut?

RJ

EMPOWERgmatRichC · Answer

Hi rjacobson99,

Unfortunately, manipulating the equation in the way that you did places the variable "t" in both pairs of parentheses, so there isn't an 'obvious' solution to maximize the value. However, in the original equation, there IS an obvious Number Property that you can use...

N(t) = -20(t - 5)^2 + 500 for 0 ≤ t ≤ 10.

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 negative term - 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

GMAT assassins aren't born, they're made, 
Rich

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=cgkwFKIZ0-M

GMATinsight · Answer

Answer: Option B

Video solution by  GMATinsight

https://www.youtube.com/watch?v=c9kzTXt2P7Q

Ceekey · Answer

I actually didn't use the "logic" approach, but rather the mathematics approach. 
So for all those who are familiar with deriving, it can be solved in 3 steps.

N=-20(t-5)²+500 it is a maximum problem, so deriving this becomes to:

0=-40(t-5) and now simply work out t.

0=-40t + 200 add 40
40t = 200 divide by 40
t = 5

or

0 =-40(t-5) divide by 40
0 = t - 5 add 5
5 = t

plus the 2 hours from the question: 5 + 2 = 7

hardikmodgil · Answer

hey just wondering , how do we know that N ( t ) needs to be solved or not . i was just doing the same thing as u guys were doing but i was also substituting N ( t ) value .

Bunuel · Answer

Check the highlighted parts in the stem. N(t) represents the depth, while we are asked to find such t for which the depth,  N(t) = -20(t - 5)^2 + 500 , reaches its maximum.

Venu01298 · Answer

Please can someone clarify these?

1. Technically isn't the correct answer 6:00 AM? I know its not part of the answer choices but (4-5=(-1)^2 = is still 480m at 6:00AM, at which point it reaches the 480?

2. The question doesn't appear to restrict (t) to integer value, which means hours and minutes can technically be considered as well. My question is if one of the answer choices were 6:30 AM instead, that'd be the right answer? Squaring something between 0 to 1 will make the value smaller, resulting in (t-5)^2 to be smaller and hence higher depth?

Thank you!

Bunuel · Answer

1. The maximum depth is at t = 5, which is 500, not 480.

2. The phrase "t hours past 2:00" implies that t must be positive.

3. t does not necessarily need to be an integer. t turns out to be integer because  500-20(t-5)^2  reaches its maximum when  -20(t-5)^2=0 , thus when  t=5 .

Please review the solutions above for more.

bhanu29 · Answer

I know it's not in the syllabus but using differentiation would be very useful(and quick) in such questions.

\(N(t) = -20(t-5)^2 + 500 \)
\(\frac{dN(t)}{dt} = -20*2*(t-5) \)

For maxima or minima or inflation point first derivative is 0 => \(-20*2*(t-5) = 0 => t=5 \)

For maxima double derivative at t=5 should be < 0
\(\frac{d^2N(t)}{dt} = -20*2 = -40 \) which is \( < 0 \)

Therefore the value of t = 5.

Question says t hours past 2:00

That implies answer is 5:00 + 2:00 = 7:00

Answer: Option B

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

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