# 03 Maximum Revenue for Tour Bus of 80 Seats

**Problem**

A tour bus has 80 seats. Experience shows that when a tour costs P28,000, all seats on the bus will be sold. For each additional P1,000 charged, however, 2 fewer seats will be sold. Find the largest possible revenue.

A. P29,000

B. P28,500

C. P28,900

D. P28,700

**Solution**

Original cost per seat

= 28,000 ÷ 80 = P350Cost per seat = 350 + 12.5(6) = P425

Number of seats sold = 80 - 2(6) = 68

Maximum Revenue = 425(68) = P28,900 ← [ C ]

= 28,000 ÷ 80 = P350

Additional cost per seat due to P1,000 increase

= 1,000 ÷ 80 = P12.5

Let *x* = number of P1,000 additional charges

Additional cost per seat = 12.5*x*

Cost per seat = 350 + 12.5*x*

Number of seats sold = 80 - 2*x*

Revenue, *R* = cost per seat × number of seats sold

$R = (350 + 12.5x)(80 - 2x)$

$R = 28,000 + 300x - 25x^2$

$\dfrac{dR}{dx} = 300 - 50x = 0$

$x = 6$

For Maximum Revenue

Number of seats sold = 80 - 2(6) = 68

Maximum Revenue = 425(68) = P28,900 ← [ C ]

*answer*

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