\n| 10<\/td>\n | [latex]400\\left(10\\right)-1600=2400[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\nThat still doesn\u2019t explain how to figure out the break-even point, the number of bikes for which the revenue equals the costs.\n\nOne way is to set the functions equal to one another and solve for [latex]x[\/latex].\n [latex]R\\left(x\\right)=C\\left(x\\right)[\/latex]<\/p>\n [latex]600x=200x+1600[\/latex]<\/p>\n [latex]400x=1600[\/latex]<\/p>\n [latex]x=4[\/latex]<\/p>\nThat means that if you sell 4 bikes, you will take in the same amount you spent to make the bikes. So selling less than 4 bikes will result in a loss and selling more than 4 bikes will result in a profit.\n\nAnother method for finding the break-even point is by graphing the two functions. You can use whichever method of graphing you find useful.\n\nTo plot points, for example, make a list of values for each function. Then use them to determine coordinates that you can plot.\n\n[latex]R\\left(x\\right)=600x[\/latex]\n \n\n\n| [latex]x[\/latex]<\/td>\n | [latex]R(x)[\/latex]<\/td>\n | [latex](x, R(x))[\/latex]<\/td>\n<\/tr>\n | \n| 0<\/td>\n | 0<\/td>\n | (0, 0)<\/td>\n<\/tr>\n | \n| 2<\/td>\n | 1200<\/td>\n | (2, 1200)<\/td>\n<\/tr>\n | \n| 4<\/td>\n | 2400<\/td>\n | (4, 2400)<\/td>\n<\/tr>\n | \n| 6<\/td>\n | 3600<\/td>\n | (6, 3600)<\/td>\n<\/tr>\n | \n| 8<\/td>\n | 4800<\/td>\n | (8, 4800)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \n\n[latex]C\\left(x\\right)=200x+1600[\/latex]\n\n \n\n\n| [latex]x[\/latex]<\/td>\n | [latex]C(x)[\/latex]<\/td>\n | [latex](x, C(x))[\/latex]<\/td>\n<\/tr>\n | \n| 0<\/td>\n | 1600<\/td>\n | (0, 1600)<\/td>\n<\/tr>\n | \n| 2<\/td>\n | 2000<\/td>\n | (2, 2000)<\/td>\n<\/tr>\n | \n| 4<\/td>\n | 2400<\/td>\n | (4, 2400)<\/td>\n<\/tr>\n | \n| 6<\/td>\n | 2800<\/td>\n | (6, 2800)<\/td>\n<\/tr>\n | \n| 8<\/td>\n | 3200<\/td>\n | (8, 3200)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\nTo use the slope and y-intercept method, first find the y-intercept by setting x = 0. Then determine the slope as the coefficient of the variable, which is the m value.\n\n \n\n\n| [latex]R(x)[\/latex]<\/td>\n | [latex]C(x)[\/latex]<\/td>\n<\/tr>\n | \n| y-intercept = 0<\/td>\n | y-intercept = 1600<\/td>\n<\/tr>\n | \n| slope = 600<\/td>\n | slope = 200<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\nUsing this information, the functions can be graphed\n\n![\"Graph](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1746\/2017\/04\/19191340\/breakeven_point-253x300.png\") \n\nNow we can see the break-even point right away. It is the point at which the cost function intersects the revenue function. It occurs when [latex]x=4[\/latex]. At this point you neither make a profit nor incur a loss.\n\nThe last question to consider is how changing your price affects your profit. If your costs remain the same, increasing the price you charge will shift your break-even point to a lower number of bikes and increase your revenue for every value of [latex]x[\/latex]. However, people may not buy as many bikes. Lowering your price will shift your break-even point to a higher number of bikes and decrease your revenue for every value of [latex]x[\/latex]. However, more people may buy your bikes.\n\nLet\u2019s take a look at the graphs for two other possible prices.\n\n![\"Graph](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1746\/2017\/04\/19191342\/price400-255x300.png\") ![\"Graph](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1746\/2017\/04\/19191344\/price1000-252x300.png\") \n\nGraphing lets you quickly see a visual representation of the functions and how they are related to one another. Notice how the break-even point shifts to 8 bikes for a lower price and only 2 bikes for a higher price.\n\nSo writing, graphing, and comparing linear functions can be quite useful. As for deciding what price people are willing to pay for your bike, that\u2019s a whole different topic!\n","rendered":"At the start of this module, you were wondering whether you could earn a profit by making bikes and were given a profit function.<\/p>\n [latex]P\\left(x\\right)=R\\left(x\\right)-C\\left(x\\right)[\/latex]<\/p>\n where<\/p>\n [latex]x[\/latex] = the number of bikes produced and sold<\/p>\n [latex]P(x)[\/latex] = profit as a function of x<\/p>\n [latex]R(x)[\/latex] = revenue as a function of x<\/p>\n [latex]C(x)[\/latex] = cost as a function of x<\/p>\n Profit, revenue, and cost are all linear functions. They are a function of the number of bikes sold. Remember that you were planning to sell each bike for $600 and it cost you $1,600 for fixed costs plus $200 per bike.<\/p>\n Since you take in $600 for each bike you sell, revenue is:<\/p>\n [latex]R\\left(x\\right)=600x[\/latex]<\/p>\n Your costs are $200 per bike plus a fixed cost of $1600, so your overall cost is:<\/p>\n [latex]C\\left(x\\right)=200x+1600[\/latex]<\/p>\n That means that profit becomes:<\/p>\n [latex]P\\left(x\\right)=600x-\\left(200x+1600\\right)[\/latex]<\/p>\n [latex]=400x-1600[\/latex]<\/p>\n With that in mind, let\u2019s take another look at the table. The profit is found by subtracting the cost function from the revenue function for each number of bikes.<\/p>\n \n \n\n\n| Number of bikes ([latex]x[\/latex])<\/td>\n | Profit ($)<\/td>\n<\/tr>\n | \n| 2<\/td>\n | [latex]400\\left(2\\right)-1600=-800[\/latex] (loss)<\/td>\n<\/tr>\n | \n| 5<\/td>\n | [latex]400\\left(5\\right)-1600=400[\/latex]<\/td>\n<\/tr>\n | \n| 10<\/td>\n | [latex]400\\left(10\\right)-1600=2400[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n That still doesn\u2019t explain how to figure out the break-even point, the number of bikes for which the revenue equals the costs.<\/p>\n One way is to set the functions equal to one another and solve for [latex]x[\/latex].<\/p>\n [latex]R\\left(x\\right)=C\\left(x\\right)[\/latex]<\/p>\n [latex]600x=200x+1600[\/latex]<\/p>\n [latex]400x=1600[\/latex]<\/p>\n [latex]x=4[\/latex]<\/p>\n That means that if you sell 4 bikes, you will take in the same amount you spent to make the bikes. So selling less than 4 bikes will result in a loss and selling more than 4 bikes will result in a profit.<\/p>\n Another method for finding the break-even point is by graphing the two functions. You can use whichever method of graphing you find useful.<\/p>\n To plot points, for example, make a list of values for each function. Then use them to determine coordinates that you can plot.<\/p>\n [latex]R\\left(x\\right)=600x[\/latex]<\/p>\n \n\n\n| [latex]x[\/latex]<\/td>\n | [latex]R(x)[\/latex]<\/td>\n | [latex](x, R(x))[\/latex]<\/td>\n<\/tr>\n | \n| 0<\/td>\n | 0<\/td>\n | (0, 0)<\/td>\n<\/tr>\n | \n| 2<\/td>\n | 1200<\/td>\n | (2, 1200)<\/td>\n<\/tr>\n | \n| 4<\/td>\n | 2400<\/td>\n | (4, 2400)<\/td>\n<\/tr>\n | \n| 6<\/td>\n | 3600<\/td>\n | (6, 3600)<\/td>\n<\/tr>\n | \n| 8<\/td>\n | 4800<\/td>\n | (8, 4800)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n <\/p>\n [latex]C\\left(x\\right)=200x+1600[\/latex]<\/p>\n \n \n\n\n| [latex]x[\/latex]<\/td>\n | [latex]C(x)[\/latex]<\/td>\n | [latex](x, C(x))[\/latex]<\/td>\n<\/tr>\n | \n| 0<\/td>\n | 1600<\/td>\n | (0, 1600)<\/td>\n<\/tr>\n | \n| 2<\/td>\n | 2000<\/td>\n | (2, 2000)<\/td>\n<\/tr>\n | \n| 4<\/td>\n | 2400<\/td>\n | (4, 2400)<\/td>\n<\/tr>\n | \n| 6<\/td>\n | 2800<\/td>\n | (6, 2800)<\/td>\n<\/tr>\n | \n| 8<\/td>\n | 3200<\/td>\n | (8, 3200)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n To use the slope and y-intercept method, first find the y-intercept by setting x = 0. Then determine the slope as the coefficient of the variable, which is the m value.<\/p>\n \n \n\n\n| [latex]R(x)[\/latex]<\/td>\n | [latex]C(x)[\/latex]<\/td>\n<\/tr>\n | \n| y-intercept = 0<\/td>\n | y-intercept = 1600<\/td>\n<\/tr>\n | \n| slope = 600<\/td>\n | slope = 200<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n Using this information, the functions can be graphed<\/p>\n <\/p>\n Now we can see the break-even point right away. It is the point at which the cost function intersects the revenue function. It occurs when [latex]x=4[\/latex]. At this point you neither make a profit nor incur a loss.<\/p>\n The last question to consider is how changing your price affects your profit. If your costs remain the same, increasing the price you charge will shift your break-even point to a lower number of bikes and increase your revenue for every value of [latex]x[\/latex]. However, people may not buy as many bikes. Lowering your price will shift your break-even point to a higher number of bikes and decrease your revenue for every value of [latex]x[\/latex]. However, more people may buy your bikes.<\/p>\n Let\u2019s take a look at the graphs for two other possible prices.<\/p>\n <\/p>\n Graphing lets you quickly see a visual representation of the functions and how they are related to one another. Notice how the break-even point shifts to 8 bikes for a lower price and only 2 bikes for a higher price.<\/p>\n So writing, graphing, and comparing linear functions can be quite useful. As for deciding what price people are willing to pay for your bike, that\u2019s a whole different topic!<\/p>\n\n\t\t\t | | | | | | | |