value of S when passenger is maximum

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by diebeatsthegmat » Tue Nov 02, 2010 7:56 pm
kushal.adhia wrote:@diebeatsthegmat

The correct statement, as u have mentioned, is p = -2(s-4)^2+32

I came across the same question when i took one of the tests...
Can u please explain how u came up with an answer for this question?

Thanks

Kushal
yes, sure,
we have to find S in order P maximum,
(s-4)^2 is always positive, so we have to find the least number of S inorder P gets max,
from all 5 answer choice, only s=4 is fit to what we need
if s=0 p=-2( 4-4)^2+32=0+32=32
( when you try another answer choice, all will give you the smaller number which is < 32) so S=4 is our answer

However if p=-2(s-4)2+32, the answer choice will be different.
in that case, s must be the smallest to get P maximum, s will be negative or get the smallest number of psitive number.
see in the 5 answers choices, we see only 1 is fit,
p=-2(1-4)*2+32=12+32

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by kushal.adhia » Wed Nov 03, 2010 11:02 pm
got it

thanks

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by fskilnik@GMATH » Thu Nov 04, 2010 4:19 am
selango wrote:First I got confused seeing the equation.

I think its (s-4)^2

P = -2(S - 4)^2 + 32

When u sub other values we need to add 32 with one negative value which ll decrease P.

Only S=4 adds 32 with 0 and gives P=32 which is high.

Pick C
Hi there, guys.

I agree, selango must have guessed the proper expression on the question stem.

Let me justify the answer (assuming the modification he mentioned)!

P = -2(S-4)^2 +32 will be P-maximized whenever the expression -2(S-4)^2 is maximized (32 is just a constant...), therefore please realize that -2(S-4)^2 is zero when S equals to 4 and NEGATIVE in all other cases... Done! ;)

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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