a hall charges $30 per person for a sports banquet when 200 people attend. for every 10 extra people that attend, the hall will decrease the price by $1.50 per person. what number of people will maximize the revenue for the hall?

Answer 1

does anyone know the answer ?

Answer 2

Let x be the number of 10-person increases. So, since revenue is price*people,

R(x) = (200+10x)(30-1.50x)

That is just a parabola, with its vertex midway between the roots, at

x = (-20+20)/2 = 0

That is, max revenue for just 200 people.

That is assuming that the price decrease applies to all the diners.

If the discount only applies to the extra people, then there is no maximum -- the more diners, the more revenue.

Answer 3

Let x be the number of people attends.

Let the revenue be y.

Let the initial revenue when 120 people attend is 30*120 = 3600

y= people attend * price

y= (120+x)*(30- 1.5*(x/10)]

y= (120+x) * ( 30 - 0.15x)

y= ( 3600 - 18x +30x - 0.15x^2)

==> y= -0.15x^2 +12x +3600

Now we need to find the maximum values.

First we will determine y'.

==> y' = -0.3x +12

Now we will find the critical values.

==> -0.3x +12 = 0

==> -0.3x = -12

==> x = -12/-0.3 = 40

The number of people that will maximize the revenue is 40 plus the initial 120 people = 40+120 = 160 people