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...

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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**