A motel has 20 rooms. If the manager charges $60 per room per night, all the rooms will be rented. For each $5 increase, one less room will be rented. How much rent should be charged to maximize revenue? What is the maximum revenue?

Rent: $80

Revenue: $1280

Step-by-step explanation:

We can solve this by representing the problem as an expression and finding the maximum of that

We know that revenue = number of sales * average price. If x represents the amount of 5 dollar increases, then we subtract x from 20 to get the total number of sales. For example, if there was 1 5 dollar increase, the number of sales would be 20-1 = 19. Similarly, the average price is equal to 60 + 5x, as we add $5 for each x. For example, if there were 3 $5 increases, it would cost 60 + 3 * 5 = 75 per room.

Therefore,

revenue = (20-x) * (60 + 5x)

= 1200 + 100x - 60x - 5x²

= -5x²+40x+1200

This is a quadratic expression -- we know it is one because it is of form ax²+bx+c, where a=-5, b=40, and c = 1200. Because the a is negative, the quadratic expression (which forms a parabola) opens downward, meaning that there is a peak of the expression at its vertex.

The x value of the vertex of a parabola given this form for the quadratic expression is (-b/2a) , which is (-40/(2*(-5))) = -40/-10 = 4 here, and plugging that into our equation,

y= -5x²+40x+1200

= -5 (4)²+40(4) + 1200

= -5 * 16 + 160 + 1200

= 1280 as our maximum revenue

Therefore, the amount of 5 dollar increases is 4, making the average price equal to 60 + 5 * 4 = 80 dollars

207/8 is the right