Optimizing Restaurant Seats for Maximum Profit

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Homework Help Overview

The discussion revolves around optimizing the number of seats in a restaurant to maximize profit, focusing on the profit per seat based on the number of seats available. The problem involves understanding profit functions and their behavior as the number of seats changes.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants suggest creating a hand plot and using calculus to find the maximum profit, while others question the necessity of calculus given the marginal profit information. There is also discussion about the profit function's definition for fewer than 60 seats and the implications of adding more seats beyond 80.

Discussion Status

Participants are exploring different methods to approach the problem, including graphical analysis and calculus. Some have provided insights into the profit function and its derivatives, while others are questioning the assumptions made regarding profit at lower seat counts. There is no explicit consensus on the best approach yet.

Contextual Notes

There is a lack of clarity regarding the profit function for fewer than 60 seats, and assumptions about marginal profit are being discussed. The problem also involves discrete considerations in profit calculation as the number of seats increases beyond 80.

coolant
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In planning a resturant, it is estimated that a profit of $5 per seat will be made if the number of seats is between 60 and 80, inclusive. On the other hand, the profit on each seat will decrease by 5 cents for each seat above 80?

a) Find the number of seats that will produce the max profit
b) what is the max profit?
 
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Make a hand plot first. What is the answer? Then use calculus to find the maximum. What is the answer? They should match, eh? Welcome to PF!
 
You don't really need calculus since the marginal profit is almost given to you. Just stop adding seats when the marginal profit is less than 0.
 
You don't say what the "profit" is, if any, if the number of seats is less than 60 so I will assume it is not defined for x< 60.

Okay, the profit function, for x seats, is
P(x)= 5x if 60<= x<= 80
= (5- 0.05(x-80))x if x> 80

As Orthodontist said, you don't really need calculus since the marginal profit is "almost given to you" but you can:

P'(x)= 5 if 60<= x<= 80
= -0.05x+ (5- 0.05(x-80))

For what value of x is that equal to 0?
 
Actually, I think the additional profit on placing the x'th seat for x > 80 is
-.05(x-1)+(5-.05(x-80))
Since with each seat, you
1. Lose 5 cents on each seat already placed, which is x-1 seats
2. Gain (5-.05(x-80)) dollars on the current seat
A strict derivative would only be approximate since this is discrete.
 

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