A health club currently charges its 1500 members monthly membership dues of $41. The board decides to increase the amount of dues charged. The marketing dept. shows that each increase of $4 will result in the loss of 4 clients. How much should the club charge to optimize the revenue from monthly dues?
Oops, each increase of $1 will result in the loss of 4 members
this is approximately the same as the other one.
did you attempt this at all?
I got $51.25 but it says that isn't right
lets see. they will make \[(41+x)(1500-4x)\] yes. where x is the dollar increase
otherwise known as \[-4 x^2+1336 x+61500\]
a parabola facing down, max is at vertex, compute \[\frac{-b}{2a}\] to get your answer
\[\frac{1336}{8}\] dang i get $167 maybe i made a mistake
increase of one dollar gives decrease of 4 members yes? so if you start at 41 and increase by x dollars you get 41+x dollars per member. decrease for every dollar is 4 members. so if you have 1500 members and increase by x dollars you get 1500-4x members. this equation looks good to me
Thank you for trying :-)
maybe i multiplied wrong hold on.
not it is right. maybe the answer really is $167
can you check?
Nope, not it.
satellite's answer looks right to me.
Sorry, I tried it and it wasn't correct.
i keep trying and it keeps looking right. are you sure you have the problem written correctly?
A health club currently charges its 1500 members monthly membership dues of $41. The board decides to increase the amount of dues charged. The marketing dept. shows that each increase of $1 will result in the loss of 4 clients. How much should the club charge to optimize the revenue from monthly dues?
hold on! satellite, did you get x = 167?
after you differentiated?
k i am holding
this is not a calc problem. vertex is \[-\frac{b}{2a}\]
Yep, double checked the problem and you wrote it correctly satellite73.
\[\frac{1336}{8}=167\]
it is the same.
then i say $167 and i am sticking with that
Ugh.
no, 167 is the amount they charge on top of 41 to get maximum profit. so they charge 167+41
oh for the love of peter. ho ho ho
all together now! facepalm!
sorry.
Hooray! That's it!
yay
Thank you both!!
you are welcome :)
also, satellite, if you differentiate a down facing parabola and equate it to 0, you will get the maxima which happens to be the vertex of the parabola. so yes, this is indeed a calculus problem. in fact, that is why the vertex of the parabola is -b/2a try it out with this problem and you will see what i'm saying.
y=ax^2+bx+c dy/dx=2ax+b 0=2ax+b -b=2ax x=-b/2a solution to what the person with the hard name issaying
dhatraditya i mean
lol, yes
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