Math Is Fun Forum

karney

Guest

cost problem

100 people will attend the theater if tickets cost $40 each.  For each $5 increase in price, 10 fewer people will attend.  What price will deliver the maximum dollar sales.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: cost problem

Hi karney;

I am asuming we cannot have more than 100 people and that the increments are 10 people and 5 dollars. Simplest way, without any math is to just enumerate all the cases. Maximum is 90 people at 45 dollars yielding 4050 dollars.

100 people * 40 bucks =4000
  90 people * 45 bucks =4050
  80 people * 50 bucks =4000
  70 people * 55 bucks =3850
  60 people * 60 bucks =3600
  50 people * 65 bucks =3250
  40 people * 70 bucks =2800
  30 people * 75 bucks =2250
  20 people * 80 bucks =1600
  10 people * 85 bucks =850

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: cost problem

Since the $40 and $50 prices give the same overall sales figure, you can tell that the $45 price will definitely give the maximum one even if fractions are allowed (because the sales figure is a quadratic function of the price).

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Why did the vector cross the road?
It wanted to be normal.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: cost problem

Hi karney;

Mathsyperson is right.

This is the formula that generates the above numbers. It is indeed a quadratic.

the derivative with respect to x is

setting to 0 and solving for x, yields x=1

So the maximum is 90 people at 45 dollars a head.

Last edited by bobbym (2009-06-01 21:49:44)

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Magister_Asmodeus

Member

Registered: 2009-05-24

Posts: 5

Re: cost problem

Hi there.

Another thing you could say is:
since your Quadratic function is f(x)=(100-10x)(40+5x) or
f(x)=-50x²+100x+4000 with a=-50, b=100, c=4000
you can say that the maximum point will be (x,y)=(-b/2a,-Δ/4a), that means x=1 and y=4050
($45 per ticket, $4050 total income)