# Pudgy Point Restaurant profits scenario

Published: 2019-12-09 07:30:00
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This scenario projects a simulation that integrates a purchase costs, purchase quantity, and sales amount integration to define the profitable combination. However, the purchase costs are consistent for particular goods. The calculations for the simulation are shown below.

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Role # 1 Drink Designer

y = 1.0 x 101 x, for apple

y = 1.2 x 101 x, for bananas

y = 1.4 x 101 x, for cherry

y = 1.6 x 101 x, for dragon fruit

Where x is the price of the respective sodas, separately, and y is the number of bottles of sodas sold separately,

Therefore

y = 10 x, for apple

y = 12 x, for bananas

y = 14 x, for cherry

y = 16 x, for dragon fruit

The graphs are shown below

This helps in finding the maximum price against the quantity of bottle of soda that are sold. This projects a multiplication of price of soda against that of bottle sold in order to find a scenario where there is maximum profit.

A graph of the equation y = 10 x, for apple

(Price of soda Cost of soda) * Bottles sold = profit

(5 - 4) * 5 = \$ 5

(6 - 4) * 4 = \$ 8

(7 - 4) * 3 = \$ 9

(8 - 4) * 2 = \$ 8

(9 - 4) * 1 = \$ 5

(10 - 4) * 0 = \$ 0

Therefore, there will be maximum profit when the price is \$ 7, which is gives a profit of \$ 9

A graph of the equation y = 12 x, for bananas

(Price of soda Cost of soda) * Bottles sold = profit

(4 - 3) * 8 = \$ 8

(5 - 3) * 7 = \$ 14

(6 - 3) * 6 = \$ 18

(7 - 3) * 5 = \$ 20

(8 - 3) * 4 = \$ 20

(9 - 3) * 3 = \$ 18

The Maximum Profit is achieved in two scenarios, where the price of soda is set at \$ 7 and \$ 8, with both \$ 20 profit.

A graph of the equation y = 14 x, for cherry

(Price of soda Cost of soda) * Bottles sold = profit

(3 2) * 11 = \$ 11

(4 2) * 10 = \$ 20

(5 2) * 9 = \$ 27

(6 2) * 8 = \$ 32

(7 2) * 7 = \$ 35

(8 2) * 6 = \$ 36

(9 2) * 5 = \$ 35

(10 2) * 4 = \$ 32

The Maximum profit is achieved when the selling price is set at \$ 8

A graph of the equation y = 16 x, for dragon fruit

(Price of soda Cost of soda) * Bottles sold = profit

(2 - 1) * 14 = \$14

(3 - 1) * 13 = \$26

(4 - 1) * 12 = \$36

(5 - 1) * 11 = \$44

(6 - 1) * 10 = \$50

(7 - 1) * 9 = \$54

(8 - 1) * 8 = \$56

(9 - 1) * 7 = \$56

(10 - 1) * 6 = \$54

The Maximum profit is achievable when the price of soda is set at two scenarios, that is, \$ 8 and \$ 9 and gives a profit of \$ 56.

Scenario questions

Question 1

I have understood how to integrate math, into simulating a business scenario through understanding general aspects of conducting experiments in real life through mathematics.

Question 2

Based on the fact that the simulations prospect a realistic world, they will help me understand the appropriateness of independence through self-employment. I will not have the urge for being employed by an organization but rather manage a business.

Question 3

I will adopt basic error stipulations and other business scenarios like discounts and the possibility of discounts, in aspects of probability and its relation to standard deviation. Through this, I will have a more accurate projection of the scenario, because this is a possibility in business.

Question 4

I have excellent understanding of simulation, however, based on the fact that I am yet to learn more from similar situation and encounter complex scenarios, I will require more exposure to more complex scenarios in order to achieve this prospect.