Maximizing profit function given cost and demand functions












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I am given the demand function $$D(x)=10x^2 + 50x$$ and a total cost of $$C(x) = x^3 + 10x$$ where $x$ is the number of units demanded. I am asked to maximize the profit so what I did is I used the formula $$P(x) = R(x) - C(x)$$ where $R(x) = x D(x)$. Then I took the derivative of $P(x)$ and equated to 0. But I got a negative value of x more so a value less than 1 since $$P'(x) = 27x^2 + 100x -10$$ Am I right or did I do something wrong in between my process? I don't think it's logical to have a quantity which is negative and less than 1 for this kind of problem. Please help.






calculus derivatives proof-verification optimization economics







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    $begingroup$


    I am given the demand function $$D(x)=10x^2 + 50x$$ and a total cost of $$C(x) = x^3 + 10x$$ where $x$ is the number of units demanded. I am asked to maximize the profit so what I did is I used the formula $$P(x) = R(x) - C(x)$$ where $R(x) = x D(x)$. Then I took the derivative of $P(x)$ and equated to 0. But I got a negative value of x more so a value less than 1 since $$P'(x) = 27x^2 + 100x -10$$ Am I right or did I do something wrong in between my process? I don't think it's logical to have a quantity which is negative and less than 1 for this kind of problem. Please help.






    calculus derivatives proof-verification optimization economics







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I am given the demand function $$D(x)=10x^2 + 50x$$ and a total cost of $$C(x) = x^3 + 10x$$ where $x$ is the number of units demanded. I am asked to maximize the profit so what I did is I used the formula $$P(x) = R(x) - C(x)$$ where $R(x) = x D(x)$. Then I took the derivative of $P(x)$ and equated to 0. But I got a negative value of x more so a value less than 1 since $$P'(x) = 27x^2 + 100x -10$$ Am I right or did I do something wrong in between my process? I don't think it's logical to have a quantity which is negative and less than 1 for this kind of problem. Please help.






      calculus derivatives proof-verification optimization economics







      share|cite|improve this question











      $endgroup$




      I am given the demand function $$D(x)=10x^2 + 50x$$ and a total cost of $$C(x) = x^3 + 10x$$ where $x$ is the number of units demanded. I am asked to maximize the profit so what I did is I used the formula $$P(x) = R(x) - C(x)$$ where $R(x) = x D(x)$. Then I took the derivative of $P(x)$ and equated to 0. But I got a negative value of x more so a value less than 1 since $$P'(x) = 27x^2 + 100x -10$$ Am I right or did I do something wrong in between my process? I don't think it's logical to have a quantity which is negative and less than 1 for this kind of problem. Please help.






      calculus derivatives proof-verification optimization economics




      calculus derivatives proof-verification optimization economics






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      share|cite|improve this question













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      edited Dec 18 '18 at 9:43









      Batominovski

      33.1k33293




      33.1k33293










      asked May 24 '15 at 8:54









      marg_ocruzmarg_ocruz

      301211




      301211






















          1 Answer
          1






          active

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          1












          $begingroup$

          You should multiply $D(x)$ by the price of a single unit, to get the total revenue.



          So if $Q$ is the price of a single unit (which presumably does not depend on $x$), you get



          $$P(x) = QD(x) - C(x)$$



          After that you are correct,to maximize this value with respect to $x$ you take the derivative and equate it to $0$ (you should show that this is indeed a maximum, and not a minimum or a stationary point. But it's not difficult to do so)






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            how do i get $Q$? I am not given the value of $Q$... :(
            $endgroup$
            – marg_ocruz
            May 24 '15 at 9:13










          • $begingroup$
            @marg_ocruz You should probably explain your question better.. Because you wrote that $x$ is the "number of units demanded" and that $D(x)$ is the "demand function".. What is the difference? It's not clear to me :-)
            $endgroup$
            – Ant
            May 24 '15 at 9:15










          • $begingroup$
            @marg_ocruz If instead $x$ is the price of a unit (which makes sense, as both $D$ is going to depend on $x$) then your calculations are correct. In this case that value you get implies setting the price to something much less than $1$.. Irrealistic, but this is just a paper problem :-)
            $endgroup$
            – Ant
            May 24 '15 at 9:16










          • $begingroup$
            the problem states that x is the number of units demanded and that D(x) is the demand function. I think it's mistakenly written by my professor.. I got your point and thanks a lot.. :)
            $endgroup$
            – marg_ocruz
            May 24 '15 at 9:21











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          1 Answer
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          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          You should multiply $D(x)$ by the price of a single unit, to get the total revenue.



          So if $Q$ is the price of a single unit (which presumably does not depend on $x$), you get



          $$P(x) = QD(x) - C(x)$$



          After that you are correct,to maximize this value with respect to $x$ you take the derivative and equate it to $0$ (you should show that this is indeed a maximum, and not a minimum or a stationary point. But it's not difficult to do so)






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            how do i get $Q$? I am not given the value of $Q$... :(
            $endgroup$
            – marg_ocruz
            May 24 '15 at 9:13










          • $begingroup$
            @marg_ocruz You should probably explain your question better.. Because you wrote that $x$ is the "number of units demanded" and that $D(x)$ is the "demand function".. What is the difference? It's not clear to me :-)
            $endgroup$
            – Ant
            May 24 '15 at 9:15










          • $begingroup$
            @marg_ocruz If instead $x$ is the price of a unit (which makes sense, as both $D$ is going to depend on $x$) then your calculations are correct. In this case that value you get implies setting the price to something much less than $1$.. Irrealistic, but this is just a paper problem :-)
            $endgroup$
            – Ant
            May 24 '15 at 9:16










          • $begingroup$
            the problem states that x is the number of units demanded and that D(x) is the demand function. I think it's mistakenly written by my professor.. I got your point and thanks a lot.. :)
            $endgroup$
            – marg_ocruz
            May 24 '15 at 9:21
















          1












          $begingroup$

          You should multiply $D(x)$ by the price of a single unit, to get the total revenue.



          So if $Q$ is the price of a single unit (which presumably does not depend on $x$), you get



          $$P(x) = QD(x) - C(x)$$



          After that you are correct,to maximize this value with respect to $x$ you take the derivative and equate it to $0$ (you should show that this is indeed a maximum, and not a minimum or a stationary point. But it's not difficult to do so)






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            how do i get $Q$? I am not given the value of $Q$... :(
            $endgroup$
            – marg_ocruz
            May 24 '15 at 9:13










          • $begingroup$
            @marg_ocruz You should probably explain your question better.. Because you wrote that $x$ is the "number of units demanded" and that $D(x)$ is the "demand function".. What is the difference? It's not clear to me :-)
            $endgroup$
            – Ant
            May 24 '15 at 9:15










          • $begingroup$
            @marg_ocruz If instead $x$ is the price of a unit (which makes sense, as both $D$ is going to depend on $x$) then your calculations are correct. In this case that value you get implies setting the price to something much less than $1$.. Irrealistic, but this is just a paper problem :-)
            $endgroup$
            – Ant
            May 24 '15 at 9:16










          • $begingroup$
            the problem states that x is the number of units demanded and that D(x) is the demand function. I think it's mistakenly written by my professor.. I got your point and thanks a lot.. :)
            $endgroup$
            – marg_ocruz
            May 24 '15 at 9:21














          1












          1








          1





          $begingroup$

          You should multiply $D(x)$ by the price of a single unit, to get the total revenue.



          So if $Q$ is the price of a single unit (which presumably does not depend on $x$), you get



          $$P(x) = QD(x) - C(x)$$



          After that you are correct,to maximize this value with respect to $x$ you take the derivative and equate it to $0$ (you should show that this is indeed a maximum, and not a minimum or a stationary point. But it's not difficult to do so)






          share|cite|improve this answer









          $endgroup$



          You should multiply $D(x)$ by the price of a single unit, to get the total revenue.



          So if $Q$ is the price of a single unit (which presumably does not depend on $x$), you get



          $$P(x) = QD(x) - C(x)$$



          After that you are correct,to maximize this value with respect to $x$ you take the derivative and equate it to $0$ (you should show that this is indeed a maximum, and not a minimum or a stationary point. But it's not difficult to do so)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 24 '15 at 9:01









          AntAnt

          17.4k22874




          17.4k22874












          • $begingroup$
            how do i get $Q$? I am not given the value of $Q$... :(
            $endgroup$
            – marg_ocruz
            May 24 '15 at 9:13










          • $begingroup$
            @marg_ocruz You should probably explain your question better.. Because you wrote that $x$ is the "number of units demanded" and that $D(x)$ is the "demand function".. What is the difference? It's not clear to me :-)
            $endgroup$
            – Ant
            May 24 '15 at 9:15










          • $begingroup$
            @marg_ocruz If instead $x$ is the price of a unit (which makes sense, as both $D$ is going to depend on $x$) then your calculations are correct. In this case that value you get implies setting the price to something much less than $1$.. Irrealistic, but this is just a paper problem :-)
            $endgroup$
            – Ant
            May 24 '15 at 9:16










          • $begingroup$
            the problem states that x is the number of units demanded and that D(x) is the demand function. I think it's mistakenly written by my professor.. I got your point and thanks a lot.. :)
            $endgroup$
            – marg_ocruz
            May 24 '15 at 9:21


















          • $begingroup$
            how do i get $Q$? I am not given the value of $Q$... :(
            $endgroup$
            – marg_ocruz
            May 24 '15 at 9:13










          • $begingroup$
            @marg_ocruz You should probably explain your question better.. Because you wrote that $x$ is the "number of units demanded" and that $D(x)$ is the "demand function".. What is the difference? It's not clear to me :-)
            $endgroup$
            – Ant
            May 24 '15 at 9:15










          • $begingroup$
            @marg_ocruz If instead $x$ is the price of a unit (which makes sense, as both $D$ is going to depend on $x$) then your calculations are correct. In this case that value you get implies setting the price to something much less than $1$.. Irrealistic, but this is just a paper problem :-)
            $endgroup$
            – Ant
            May 24 '15 at 9:16










          • $begingroup$
            the problem states that x is the number of units demanded and that D(x) is the demand function. I think it's mistakenly written by my professor.. I got your point and thanks a lot.. :)
            $endgroup$
            – marg_ocruz
            May 24 '15 at 9:21
















          $begingroup$
          how do i get $Q$? I am not given the value of $Q$... :(
          $endgroup$
          – marg_ocruz
          May 24 '15 at 9:13




          $begingroup$
          how do i get $Q$? I am not given the value of $Q$... :(
          $endgroup$
          – marg_ocruz
          May 24 '15 at 9:13












          $begingroup$
          @marg_ocruz You should probably explain your question better.. Because you wrote that $x$ is the "number of units demanded" and that $D(x)$ is the "demand function".. What is the difference? It's not clear to me :-)
          $endgroup$
          – Ant
          May 24 '15 at 9:15




          $begingroup$
          @marg_ocruz You should probably explain your question better.. Because you wrote that $x$ is the "number of units demanded" and that $D(x)$ is the "demand function".. What is the difference? It's not clear to me :-)
          $endgroup$
          – Ant
          May 24 '15 at 9:15












          $begingroup$
          @marg_ocruz If instead $x$ is the price of a unit (which makes sense, as both $D$ is going to depend on $x$) then your calculations are correct. In this case that value you get implies setting the price to something much less than $1$.. Irrealistic, but this is just a paper problem :-)
          $endgroup$
          – Ant
          May 24 '15 at 9:16




          $begingroup$
          @marg_ocruz If instead $x$ is the price of a unit (which makes sense, as both $D$ is going to depend on $x$) then your calculations are correct. In this case that value you get implies setting the price to something much less than $1$.. Irrealistic, but this is just a paper problem :-)
          $endgroup$
          – Ant
          May 24 '15 at 9:16












          $begingroup$
          the problem states that x is the number of units demanded and that D(x) is the demand function. I think it's mistakenly written by my professor.. I got your point and thanks a lot.. :)
          $endgroup$
          – marg_ocruz
          May 24 '15 at 9:21




          $begingroup$
          the problem states that x is the number of units demanded and that D(x) is the demand function. I think it's mistakenly written by my professor.. I got your point and thanks a lot.. :)
          $endgroup$
          – marg_ocruz
          May 24 '15 at 9:21


















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