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In planning a restaurant, it is estimated that a profit of $10 per seat will be made if the number of seats is no more than 40 inclusive. On the other hand, the profit on each seat will decrease by 20 cents for each seat above 40.

a)Find the number of seats that will produce the maximum profit.

b) What is the maximum profit?

Step-by-step explanation:

Let x represent the seating capacity

Number of seats = 40+x

Profit per seat = 10 - 0.20x

For maximum number of seats

P(x) = ( 40+x ) ( 10-0.20x )

P(x) = 400+10x-8x-0.2x^2

P(x) = 400+2x- 0.2x^2

Differentiating with respect to ( x )

= 2 - 0.4x

0.4x = 2

x = 2/0.4

x = 5

The seating capacity will be 40+5 = 45

For the maximum profits

40X10+ 9.9 + 9.8 + 9.7 + 9.6 + 9.5 + 9.4 + 9.3 + 9.2 + 9.1 + ... 1.0, 0.9, 0.8, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1

= 400 + an arithmetic series (first term = 0.1, common difference = 0.1, number of terms = 8+ 40 = 48 )

= 400 + (48/2)(2X0.1 + (48-1)X0.1)

= 400 + 24(0.2 + 4.7)

= 400 + 24(4.9)

= 400 + 117.6

= 517.6

= 517.6dollars

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