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by Newaz111 » Mon May 04, 2015 7:29 am
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
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by GMATGuruNY » Mon May 04, 2015 7:39 am
The problem should read as follows:
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
Depth = 500 - 20(t-5)².

To MAXIMIZE the depth, we must MINIMIZE the value in red.
Since the square of a value cannot be negative, (t-5)² ≥ 0.
Thus, the value in red will be minimized when (t-5)² = 0.
Since (t-5)² = 0 when t=5, the depth will be at its maximum 5 hours after 2am:
2am + 5 hours = 7am.

The correct answer is B.
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by [email protected] » Mon May 04, 2015 11:57 am
Hi Newaz111,

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

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