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MGMAT cat 6B PS

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MGMAT cat 6B PS

by vkb16 » Wed Apr 22, 2009 6:57 am
The number of passengers on a certain bus at any given time is given by the equation P = –2(S – 4)2 + 32, where P is the number of passengers and S is the number of stops the bus has made since beginning its route. If the bus begins its route with no passengers, what is the value of S when the bus has its greatest number of passengers?

a. 9

b. 6

c. 4

d.2

e.1

oa is C
I kinda solved my way through it, but can some one explain me the logic?
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by vkb16 » Wed Apr 22, 2009 6:59 am
mgmat says

The bus will carry its greatest passenger load when P is at its maximum value. If P = –2(S – 4)2 + 32, the maximum value of P is 32 because (S – 4)2 will never be negative, so the expression –2(S – 4)2 will never be positive. The maximum value for P will occur when –2(S – 4)2 = 0, i.e. when S = 4.

but I cant get the logic why ''the maximum value of P is 32''
:roll:
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by moutar » Wed Apr 22, 2009 10:58 am
I differentiated to check. Obviously it can be done by logic. I'll write both.

Differentiating:

Max when dP/dS = 0

dP/dS = 2(S-4) (by the chain rule)

2(S-4) = 0 when S = 4


Logic:

As it says (S-4)^2 is always positive (or 0).

So -2(S-4)^2 is always negative and will take passengers away from the 32. The largest number -2(S-4)^2 can be is zero i.e when [spoiler]S=4[/spoiler]
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