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The depth D(t), in meters, of the certain cave at t hours past 4:00

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The depth D(t), in meters, of the certain cave at t hours past 4:00 [#permalink] New post   11 Mar 2021, 23:29
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The depth D(t), in meters, of the certain cave at t hours past 4:00 in the morning is given by D(t)=-30\((t-3)^2\)+300 for 0<t<8. According to function, at what time in the morning is the depth of the cave at its maximum?

A) 4:00
B) 5:00
C) 6:00
D) 7:00
E) 8:00
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Re: The depth D(t), in meters, of the certain cave at t hours past 4:00 [#permalink] New post   11 Mar 2021, 23:31
When t = 3, the value of −30\((t − 3)^2\) + 300 is 300. So its max for all number between 0<t<8. Thus depth 3 hours after 4:00 in the morning is 7:00 in the morning.
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The depth D(t), in meters, of the certain cave at t hours past 4:00 [#permalink] New post   25 Apr 2021, 02:52
producer maximization condition
D(t)'=0 so, (-30*(t-3)^2+300)'=0 =>-60*(t-3)=0, so t=3 and
4+3=7
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Re: The depth D(t), in meters, of the certain cave at t hours past 4:00 [#permalink] New post   25 Apr 2021, 11:37
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Solution:

D(t)=-30(t−3)^2+300

(t-3)^2 is a positive value.

When multiplied with -30,it shall generate a negative number and hence we have to minimize (t-3) in order to maximize D(t).

=> t-3 is minimum at t=3

=>3 hrs past 4 = 7:00 in the morning (option d)

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Re: The depth D(t), in meters, of the certain cave at t hours past 4:00 [#permalink]
25 Apr 2021, 11:37
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