# modeling the maximum revenue

A small theatre has a seating capactiy of 2000. When the ticket price is 20, attendance is 1500> For each \$1 decrease in price, attendance increases by 100.

Write the function that represents the revenue, R(x) as a function of ticket price, x. SPecift the appropriate domain of this function.

What ticket price will yied maximum revenue? what is the maximum revenue?

### 3 個解答

• 最佳解答

A small theatre has a seating capactiy of

2000. When the ticket price is 20, attendance

is 1500> For each \$1 decrease in price, attendance increases by 100.

Write the function that represents the revenue, R(x) as a function of ticket

price, x.

SPecift the appropriate domain of this function.

What ticket price will yied maximum revenue? what is the maximum revenue?

Sol

(1) if ticket

price decrease unit is \$1

R(x)=20－x when x=0 or

1 or 2 or 3 or 4 or 5

revenue A(x)=(20－x)*(1500+100x)

So

A(0)=20*1500= 30000

A(1)=19*1600= 30400

A(2)=18*1700= 30600

A(3)=17*1800= 30600

A(4)=16*1900=30400

So

when ticket price is 18 or 17

the maximum

revenue is 30600

(2) if ticket

price decrease unit is \$0.01

R(x)=20－x when

20>=x>= 0

A(x)=(20－x)(1500+100x)

=30000+2000x－1500x－100x^2

=－100x^2+500x+30000

=－100(x^2－5x)+30000

=－100(x^2－5x+6.25)+30000+625

=－100(x－2.5)^2+30625

when x=2.5

ticket price is 17.5

the maximum

revenue is 30625

• (1) ----> 20 - X > 0

(2) ----> 1500 + 100X <= 2000

R(X) = (1500 + 100X)(20-X) = 30000 + 500X - 100X^2

= -100(X^2-5X-300) = -100[(X-5/2)^2 - 1225/4 ]

當 X = 5/2=2.5 時 Revenue , R(X) 為最大,

但是題目並未告知每 0.5 對應觀眾人數 的關係, 所以 X=2 或 3 代入

當 價格=X 下降 2元 ---> 觀眾 = 1500 + 200 = 1700 人 ---> 門票 = 18 元

Revenue = 18*1700 = 30600 元

當 價格=X 下降 3元 ---> 觀眾 = 1500 + 300 = 1800 人 ---> 門票 = 17 元

Revenue = 17*1800 = 30600 元

所以當價格 2 元 門票 = 18 元 或 價格 3 元 門票 = 17 元

老闆賺的錢都是最多 = 30600 元

• whar do u want to express?