{"id":16263,"date":"2019-10-02T20:23:09","date_gmt":"2019-10-02T20:23:09","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-applications-of-multiplying-and-dividing-polynomials\/"},"modified":"2020-10-22T09:30:33","modified_gmt":"2020-10-22T09:30:33","slug":"read-applications-of-multiplying-and-dividing-polynomials","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/chapter\/read-applications-of-multiplying-and-dividing-polynomials\/","title":{"raw":"12.4.b - Polynomials Involving Cost, Revenue, and Profit","rendered":"12.4.b – Polynomials Involving Cost, Revenue, and Profit"},"content":{"raw":"
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Learning Outcomes<\/h3>\r\n
    \r\n \t
  • Write polynomials involving cost, revenue, and profit<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn this section, we will see that polynomials are sometimes used to describe cost and revenue.<\/span>\r\n\r\n<\/div>\r\nProfit is typically defined in business as the difference between the amount of money earned (revenue) by producing a certain number of items and the amount of money it takes to produce that number of items. When you are in business, you definitely want to see profit, so it is important to know what your cost and revenue is.\r\n\r\n[caption id=\"attachment_4645\" align=\"alignleft\" width=\"300\"]\"Pile Cell Phones[\/caption]\r\n\r\nFor example, let's say that the cost to a manufacturer to produce a certain number of things is C and the revenue generated by selling those things is R. \u00a0The profit, P, can then be defined as\r\n

    P = R-C<\/p>\r\n

    The example we will work with is a hypothetical cell phone manufacturer whose\u00a0cost to manufacture x number of phones is [latex]C=2000x+750,000[\/latex], and the\u00a0Revenue generated from manufacturing x number of cell phones is [latex]R=-0.09x^2+7000x[\/latex].<\/p>\r\n\r\n

    \r\n

    Example<\/h3>\r\nDefine a Profit polynomial for the hypothetical cell phone manufacturer.\r\n[reveal-answer q=\"50187\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"50187\"]\r\n\r\nRead and Understand: <\/strong>Profit is the difference between revenue and cost, so we will need to define P = R - C for the company.\r\n\r\nDefine and Translate:<\/strong>\u00a0[latex]\\begin{array}{c}R=-0.09x^2+7000x\\\\C=2000x+750,000\\end{array}[\/latex]\r\n\r\n \r\n\r\nWrite and Solve:\u00a0<\/strong>Substitute the expressions for R and C into the Profit equation.\r\n

    [latex]\\begin{array}{c}P=R-C\\\\=-0.09x^2+7000x-\\left(2000x+750,000\\right)\\\\=-0.09x^2+7000x-2000x-750,000\\\\=-0.09x^2+5000x-750,000\\end{array}[\/latex]<\/p>\r\n

    Remember that when you subtract a polynomial, you have to subtract every term of the polynomial.<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex]P=-0.09x^2+5000x-750,000[\/latex]\r\n\r\n \r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nMathematical models are great when you use them to learn important information. \u00a0The cell phone manufacturing company can use the profit equation to find out how much profit they will make given x number of phones are manufactured. \u00a0In the next example, we will explore some profit values for this company.\r\n
    \r\n

    Example<\/h3>\r\nGiven the following numbers of cell phones manufactured, find the profit for the cell phone manufacturer:\r\n
      \r\n \t
    1. x = [latex]100[\/latex] phones<\/li>\r\n \t
    2. x = [latex]25,000[\/latex] phones<\/li>\r\n \t
    3. x= [latex]60,000[\/latex] phones<\/li>\r\n<\/ol>\r\nInterpret your results.\r\n[reveal-answer q=\"398706\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"398706\"]\r\n\r\nRead and Understand:\u00a0<\/strong>The profit polynomial defined in the previous example, [latex]P=-0.09x^2+5000x-750,000[\/latex], gives profit based on x number of phones manufactured. \u00a0We need to substitute the given numbers of phones manufactured into the equation, then try to understand what our answer means in terms of profit and number of phones manufactured.\r\n\r\nWe will move straight into write and solve since we already have our polynomial. It is probably easiest to use a calculator since the numbers in this problem are so large.\r\n\r\nWrite and Solve:\u00a0<\/strong>\r\n\r\nSubstitute x = [latex]100[\/latex]\r\n

      [latex]\\begin{array}{c}P=-0.09x^2+5000x-750,000\\\\=-0.09\\left(100\\right)^2+5000\\left(100\\right)-750,000\\\\=-900+500,000-750,000\\\\=-250,900\\end{array}[\/latex]<\/p>\r\n

      Interpret:\u00a0<\/strong>When the number of phones manufactured is [latex]100[\/latex], the profit for the business is $[latex]-250,000[\/latex]. \u00a0This is not what we want! \u00a0The company must produce more than [latex]100[\/latex] phones to make a profit.<\/p>\r\nWrite and Solve:\u00a0<\/strong>\r\n\r\nSubstitute x =[latex]25,000[\/latex]\r\n

      [latex]\\begin{array}{c}P=-0.09x^2+5000x-750,000\\\\=-0.09\\left(25000\\right)^2+5000\\left(25000\\right)-750,000\\\\=-6120000+125,000,000-750,000\\\\=118,130,000\\end{array}[\/latex]<\/p>\r\n

      Interpret:\u00a0<\/strong>When the number of phones manufactured is [latex]25,000[\/latex], the profit for the business is $[latex]118,130,000[\/latex]. \u00a0This is more like it! \u00a0If the company makes [latex]25,000[\/latex] phones it will make a profit after it pays all it's bills.<\/p>\r\n

      \u00a0If this is true, then the company should make even more phones so it can make even more money, right? \u00a0Actually, something different happens as the number of items manufactured increases without bound.<\/p>\r\nWrite and Solve:\u00a0<\/strong>\r\n\r\nSubstitute x = [latex]60,000[\/latex]\r\n

      [latex]\\begin{array}{c}P=-0.09x^2+5000x-750,000\\\\=-0.09\\left(60000\\right)^2+5000\\left(60000\\right)-750,000\\\\=-324,000,000+300,000,000-750,000\\\\=-24,750,000\\end{array}[\/latex]<\/p>\r\n

      Interpret:\u00a0<\/strong>When the number of phones manufactured is [latex]60,000[\/latex], the profit for the business is $[latex]-24,750,000[\/latex]. \u00a0Wait a minute! If the company makes [latex]60,000[\/latex] phones it will\u00a0lose money.\u00a0 What happened? At some point, the cost to manufacture the phones will overcome the amount of profit that the business can make. \u00a0If this is interesting to you, you may enjoy reading about Economics and Business models.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

      \r\n

      Try It<\/h3>\r\n[ohm_question]35200[\/ohm_question]\r\n\r\n<\/div>\r\nIn the video that follows, we present another example of finding a polynomial profit equation.\r\n\r\nhttps:\/\/youtu.be\/-TWjDC4g9dU\r\n\r\n \r\n\r\nWe have shown that profit can be modeled with a polynomial, and that the profit a company can make based on a business model like this has its bounds.","rendered":"
      \n
      \n

      Learning Outcomes<\/h3>\n
        \n
      • Write polynomials involving cost, revenue, and profit<\/li>\n<\/ul>\n<\/div>\n

        In this section, we will see that polynomials are sometimes used to describe cost and revenue.<\/span><\/p>\n<\/div>\n

        Profit is typically defined in business as the difference between the amount of money earned (revenue) by producing a certain number of items and the amount of money it takes to produce that number of items. When you are in business, you definitely want to see profit, so it is important to know what your cost and revenue is.<\/p>\n

        \"Pile<\/p>\n

        Cell Phones<\/p>\n<\/div>\n

        For example, let’s say that the cost to a manufacturer to produce a certain number of things is C and the revenue generated by selling those things is R. \u00a0The profit, P, can then be defined as<\/p>\n

        P = R-C<\/p>\n

        The example we will work with is a hypothetical cell phone manufacturer whose\u00a0cost to manufacture x number of phones is [latex]C=2000x+750,000[\/latex], and the\u00a0Revenue generated from manufacturing x number of cell phones is [latex]R=-0.09x^2+7000x[\/latex].<\/p>\n

        \n

        Example<\/h3>\n

        Define a Profit polynomial for the hypothetical cell phone manufacturer.<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        Read and Understand: <\/strong>Profit is the difference between revenue and cost, so we will need to define P = R – C for the company.<\/p>\n

        Define and Translate:<\/strong>\u00a0[latex]\\begin{array}{c}R=-0.09x^2+7000x\\\\C=2000x+750,000\\end{array}[\/latex]<\/p>\n

         <\/p>\n

        Write and Solve:\u00a0<\/strong>Substitute the expressions for R and C into the Profit equation.<\/p>\n

        [latex]\\begin{array}{c}P=R-C\\\\=-0.09x^2+7000x-\\left(2000x+750,000\\right)\\\\=-0.09x^2+7000x-2000x-750,000\\\\=-0.09x^2+5000x-750,000\\end{array}[\/latex]<\/p>\n

        Remember that when you subtract a polynomial, you have to subtract every term of the polynomial.<\/p>\n

        Answer<\/h4>\n

        [latex]P=-0.09x^2+5000x-750,000[\/latex]<\/p>\n

         <\/p>\n<\/div>\n<\/div>\n<\/div>\n

        Mathematical models are great when you use them to learn important information. \u00a0The cell phone manufacturing company can use the profit equation to find out how much profit they will make given x number of phones are manufactured. \u00a0In the next example, we will explore some profit values for this company.<\/p>\n

        \n

        Example<\/h3>\n

        Given the following numbers of cell phones manufactured, find the profit for the cell phone manufacturer:<\/p>\n

          \n
        1. x = [latex]100[\/latex] phones<\/li>\n
        2. x = [latex]25,000[\/latex] phones<\/li>\n
        3. x= [latex]60,000[\/latex] phones<\/li>\n<\/ol>\n

          Interpret your results.<\/p>\n

          Show Solution<\/span><\/p>\n
          \n

          Read and Understand:\u00a0<\/strong>The profit polynomial defined in the previous example, [latex]P=-0.09x^2+5000x-750,000[\/latex], gives profit based on x number of phones manufactured. \u00a0We need to substitute the given numbers of phones manufactured into the equation, then try to understand what our answer means in terms of profit and number of phones manufactured.<\/p>\n

          We will move straight into write and solve since we already have our polynomial. It is probably easiest to use a calculator since the numbers in this problem are so large.<\/p>\n

          Write and Solve:\u00a0<\/strong><\/p>\n

          Substitute x = [latex]100[\/latex]<\/p>\n

          [latex]\\begin{array}{c}P=-0.09x^2+5000x-750,000\\\\=-0.09\\left(100\\right)^2+5000\\left(100\\right)-750,000\\\\=-900+500,000-750,000\\\\=-250,900\\end{array}[\/latex]<\/p>\n

          Interpret:\u00a0<\/strong>When the number of phones manufactured is [latex]100[\/latex], the profit for the business is $[latex]-250,000[\/latex]. \u00a0This is not what we want! \u00a0The company must produce more than [latex]100[\/latex] phones to make a profit.<\/p>\n

          Write and Solve:\u00a0<\/strong><\/p>\n

          Substitute x =[latex]25,000[\/latex]<\/p>\n

          [latex]\\begin{array}{c}P=-0.09x^2+5000x-750,000\\\\=-0.09\\left(25000\\right)^2+5000\\left(25000\\right)-750,000\\\\=-6120000+125,000,000-750,000\\\\=118,130,000\\end{array}[\/latex]<\/p>\n

          Interpret:\u00a0<\/strong>When the number of phones manufactured is [latex]25,000[\/latex], the profit for the business is $[latex]118,130,000[\/latex]. \u00a0This is more like it! \u00a0If the company makes [latex]25,000[\/latex] phones it will make a profit after it pays all it’s bills.<\/p>\n

          \u00a0If this is true, then the company should make even more phones so it can make even more money, right? \u00a0Actually, something different happens as the number of items manufactured increases without bound.<\/p>\n

          Write and Solve:\u00a0<\/strong><\/p>\n

          Substitute x = [latex]60,000[\/latex]<\/p>\n

          [latex]\\begin{array}{c}P=-0.09x^2+5000x-750,000\\\\=-0.09\\left(60000\\right)^2+5000\\left(60000\\right)-750,000\\\\=-324,000,000+300,000,000-750,000\\\\=-24,750,000\\end{array}[\/latex]<\/p>\n

          Interpret:\u00a0<\/strong>When the number of phones manufactured is [latex]60,000[\/latex], the profit for the business is $[latex]-24,750,000[\/latex]. \u00a0Wait a minute! If the company makes [latex]60,000[\/latex] phones it will\u00a0lose money.\u00a0 What happened? At some point, the cost to manufacture the phones will overcome the amount of profit that the business can make. \u00a0If this is interesting to you, you may enjoy reading about Economics and Business models.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

          \n

          Try It<\/h3>\n