{"id":180,"date":"2023-06-21T13:22:41","date_gmt":"2023-06-21T13:22:41","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/putting-it-together-linear-and-absolute-value-equations\/"},"modified":"2023-07-09T03:22:06","modified_gmt":"2023-07-09T03:22:06","slug":"putting-it-together-linear-and-absolute-value-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/putting-it-together-linear-and-absolute-value-equations\/","title":{"raw":"Putting It Together: Linear Functions","rendered":"Putting It Together: Linear Functions"},"content":{"raw":"At the start of this module, you were wondering whether you could earn a profit by making bikes and were given a profit function.\r\n

[latex]P\\left(x\\right)=R\\left(x\\right)-C\\left(x\\right)[\/latex]<\/p>\r\n

where<\/p>\r\n

[latex]x[\/latex] = the number of bikes produced and sold<\/p>\r\n

[latex]P(x)[\/latex] = profit as a function of x<\/p>\r\n

[latex]R(x)[\/latex] = revenue as a function of x<\/p>\r\n

[latex]C(x)[\/latex] = cost as a function of x<\/p>\r\nProfit, 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.\r\n\r\nSince you take in $600 for each bike you sell, revenue is:\r\n

[latex]R\\left(x\\right)=600x[\/latex]<\/p>\r\nYour costs are $200 per bike plus a fixed cost of $1600, so your overall cost is:\r\n

[latex]C\\left(x\\right)=200x+1600[\/latex]<\/p>\r\nThat means that profit becomes:\r\n

[latex]P\\left(x\\right)=600x-\\left(200x+1600\\right)[\/latex]<\/p>\r\n

[latex]=400x-1600[\/latex]<\/p>\r\nWith 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.\r\n

\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Number of bikes ([latex]x[\/latex])<\/td>\r\nProfit ($)<\/td>\r\n<\/tr>\r\n
2<\/td>\r\n[latex]400\\left(2\\right)-1600=-800[\/latex] \u00a0(loss)<\/td>\r\n<\/tr>\r\n
5<\/td>\r\n[latex]400\\left(5\\right)-1600=400[\/latex]<\/td>\r\n<\/tr>\r\n
10<\/td>\r\n[latex]400\\left(10\\right)-1600=2400[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nThat still doesn\u2019t explain how to figure out the break-even point, the number of bikes for which the revenue equals the costs.\r\n\r\nOne way is to set the functions equal to one another and solve for [latex]x[\/latex].\r\n

[latex]R\\left(x\\right)=C\\left(x\\right)[\/latex]<\/p>\r\n

[latex]600x=200x+1600[\/latex]<\/p>\r\n

[latex]400x=1600[\/latex]<\/p>\r\n

[latex]x=4[\/latex]<\/p>\r\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.\r\n\r\nAnother method for finding the break-even point is by graphing the two functions. \u00a0You can use whichever method of graphing you find useful.\r\n\r\nTo plot points, for example, make a list of values for each function. \u00a0Then use them to determine coordinates that you can plot.\r\n\r\n[latex]R\\left(x\\right)=600x[\/latex]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
[latex]x[\/latex]<\/td>\r\n[latex]R(x)[\/latex]<\/td>\r\n[latex](x, R(x))[\/latex]<\/td>\r\n<\/tr>\r\n
0<\/td>\r\n0<\/td>\r\n(0, 0)<\/td>\r\n<\/tr>\r\n
2<\/td>\r\n1200<\/td>\r\n(2, 1200)<\/td>\r\n<\/tr>\r\n
4<\/td>\r\n2400<\/td>\r\n(4, 2400)<\/td>\r\n<\/tr>\r\n
6<\/td>\r\n3600<\/td>\r\n(6, 3600)<\/td>\r\n<\/tr>\r\n
8<\/td>\r\n4800<\/td>\r\n(8, 4800)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\n[latex]C\\left(x\\right)=200x+1600[\/latex]\r\n
\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
[latex]x[\/latex]<\/td>\r\n[latex]C(x)[\/latex]<\/td>\r\n[latex](x, C(x))[\/latex]<\/td>\r\n<\/tr>\r\n
0<\/td>\r\n1600<\/td>\r\n(0, 1600)<\/td>\r\n<\/tr>\r\n
2<\/td>\r\n2000<\/td>\r\n(2, 2000)<\/td>\r\n<\/tr>\r\n
4<\/td>\r\n2400<\/td>\r\n(4, 2400)<\/td>\r\n<\/tr>\r\n
6<\/td>\r\n2800<\/td>\r\n(6, 2800)<\/td>\r\n<\/tr>\r\n
8<\/td>\r\n3200<\/td>\r\n(8, 3200)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\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.\r\n
\r\n\r\n\r\n\r\n\r\n\r\n
\u00a0[latex]R(x)[\/latex]<\/td>\r\n\u00a0[latex]C(x)[\/latex]<\/td>\r\n<\/tr>\r\n
y-intercept = 0<\/td>\r\ny-intercept = 1600<\/td>\r\n<\/tr>\r\n
slope = 600<\/td>\r\nslope = 200<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nUsing this information, the functions can be graphed\r\n\r\n\"Graph\r\n\r\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.\r\n\r\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.\r\n\r\nLet\u2019s take a look at the graphs for two other possible prices.\r\n\r\n\"Graph\"Graph\r\n\r\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.\r\n\r\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!","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\n\n\n
Number of bikes ([latex]x[\/latex])<\/td>\nProfit ($)<\/td>\n<\/tr>\n
2<\/td>\n[latex]400\\left(2\\right)-1600=-800[\/latex] \u00a0(loss)<\/td>\n<\/tr>\n
5<\/td>\n[latex]400\\left(5\\right)-1600=400[\/latex]<\/td>\n<\/tr>\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. \u00a0You can use whichever method of graphing you find useful.<\/p>\n

To plot points, for example, make a list of values for each function. \u00a0Then use them to determine coordinates that you can plot.<\/p>\n

[latex]R\\left(x\\right)=600x[\/latex]<\/p>\n\n\n\n\n\n\n\n\n
[latex]x[\/latex]<\/td>\n[latex]R(x)[\/latex]<\/td>\n[latex](x, R(x))[\/latex]<\/td>\n<\/tr>\n
0<\/td>\n0<\/td>\n(0, 0)<\/td>\n<\/tr>\n
2<\/td>\n1200<\/td>\n(2, 1200)<\/td>\n<\/tr>\n
4<\/td>\n2400<\/td>\n(4, 2400)<\/td>\n<\/tr>\n
6<\/td>\n3600<\/td>\n(6, 3600)<\/td>\n<\/tr>\n
8<\/td>\n4800<\/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\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
0<\/td>\n1600<\/td>\n(0, 1600)<\/td>\n<\/tr>\n
2<\/td>\n2000<\/td>\n(2, 2000)<\/td>\n<\/tr>\n
4<\/td>\n2400<\/td>\n(4, 2400)<\/td>\n<\/tr>\n
6<\/td>\n2800<\/td>\n(6, 2800)<\/td>\n<\/tr>\n
8<\/td>\n3200<\/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\n\n
\u00a0[latex]R(x)[\/latex]<\/td>\n\u00a0[latex]C(x)[\/latex]<\/td>\n<\/tr>\n
y-intercept = 0<\/td>\ny-intercept = 1600<\/td>\n<\/tr>\n
slope = 600<\/td>\nslope = 200<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Using this information, the functions can be graphed<\/p>\n

\"Graph<\/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

\"Graph\"Graph<\/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

\n\t\t\t

Candela Citations<\/h3>\n\t\t\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t\t\t
CC licensed content, Original<\/div>