What (if any) system of equations would solve this problem?












1












$begingroup$


I am helping my daughter with her high school pre-calc. We've both got stuck on this. Below is a copy of the exercise as presented.




We've come up with the following equations (none of which are in the answer set).



$$ A = -y^2 + 600y $$
$$ A = -x^2 + 300x $$



Since both areas are equal:



$$ -y^2 + 600y = -x^2 + 300x $$
or
$$ y^2 - 600y = x^2 - 300x $$



Three trivial solutions (out of infinitely many): $(0,0),(300,0)$ and $(0,600)$ (curious ways to build a fence) can be seen in the graph below. Knowing the area would not make this problem seem any less strange.



Am I missing something embarrassingly obvious here?



Edit: I originally had the second equation as $ A = -2y^2 + 600y $ and I "fixed" it. Working too late and losing efficiency.



This leads to



$$-2y^2 + 600y = -x^2 + 300x $$
or
$$2(y^2 - 300y) = (x^2 - 300x) $$



which produces a similar-looking graph and trivial (zero-area) results (among infinitely many): $(0,0), (0,300)$ and $(300,0)$.



enter image description here



Question from HW






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think here the area $A$ is being treated as a known quantity. You have to select the particular $(x,y)$ that simultaneously satisfies $y^2 - 600y = x^2 - 300x,A = -x^2 + 300x,A = -y^2 + 600y$
    $endgroup$
    – Shubham Johri
    Dec 20 '18 at 7:52










  • $begingroup$
    You know what? I had -2y^2 and I "fixed" it.
    $endgroup$
    – Adam Hrankowski
    Dec 20 '18 at 8:05
















1












$begingroup$


I am helping my daughter with her high school pre-calc. We've both got stuck on this. Below is a copy of the exercise as presented.




We've come up with the following equations (none of which are in the answer set).



$$ A = -y^2 + 600y $$
$$ A = -x^2 + 300x $$



Since both areas are equal:



$$ -y^2 + 600y = -x^2 + 300x $$
or
$$ y^2 - 600y = x^2 - 300x $$



Three trivial solutions (out of infinitely many): $(0,0),(300,0)$ and $(0,600)$ (curious ways to build a fence) can be seen in the graph below. Knowing the area would not make this problem seem any less strange.



Am I missing something embarrassingly obvious here?



Edit: I originally had the second equation as $ A = -2y^2 + 600y $ and I "fixed" it. Working too late and losing efficiency.



This leads to



$$-2y^2 + 600y = -x^2 + 300x $$
or
$$2(y^2 - 300y) = (x^2 - 300x) $$



which produces a similar-looking graph and trivial (zero-area) results (among infinitely many): $(0,0), (0,300)$ and $(300,0)$.



enter image description here



Question from HW






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think here the area $A$ is being treated as a known quantity. You have to select the particular $(x,y)$ that simultaneously satisfies $y^2 - 600y = x^2 - 300x,A = -x^2 + 300x,A = -y^2 + 600y$
    $endgroup$
    – Shubham Johri
    Dec 20 '18 at 7:52










  • $begingroup$
    You know what? I had -2y^2 and I "fixed" it.
    $endgroup$
    – Adam Hrankowski
    Dec 20 '18 at 8:05














1












1








1





$begingroup$


I am helping my daughter with her high school pre-calc. We've both got stuck on this. Below is a copy of the exercise as presented.




We've come up with the following equations (none of which are in the answer set).



$$ A = -y^2 + 600y $$
$$ A = -x^2 + 300x $$



Since both areas are equal:



$$ -y^2 + 600y = -x^2 + 300x $$
or
$$ y^2 - 600y = x^2 - 300x $$



Three trivial solutions (out of infinitely many): $(0,0),(300,0)$ and $(0,600)$ (curious ways to build a fence) can be seen in the graph below. Knowing the area would not make this problem seem any less strange.



Am I missing something embarrassingly obvious here?



Edit: I originally had the second equation as $ A = -2y^2 + 600y $ and I "fixed" it. Working too late and losing efficiency.



This leads to



$$-2y^2 + 600y = -x^2 + 300x $$
or
$$2(y^2 - 300y) = (x^2 - 300x) $$



which produces a similar-looking graph and trivial (zero-area) results (among infinitely many): $(0,0), (0,300)$ and $(300,0)$.



enter image description here



Question from HW






algebra-precalculus







share|cite|improve this question











$endgroup$




I am helping my daughter with her high school pre-calc. We've both got stuck on this. Below is a copy of the exercise as presented.




We've come up with the following equations (none of which are in the answer set).



$$ A = -y^2 + 600y $$
$$ A = -x^2 + 300x $$



Since both areas are equal:



$$ -y^2 + 600y = -x^2 + 300x $$
or
$$ y^2 - 600y = x^2 - 300x $$



Three trivial solutions (out of infinitely many): $(0,0),(300,0)$ and $(0,600)$ (curious ways to build a fence) can be seen in the graph below. Knowing the area would not make this problem seem any less strange.



Am I missing something embarrassingly obvious here?



Edit: I originally had the second equation as $ A = -2y^2 + 600y $ and I "fixed" it. Working too late and losing efficiency.



This leads to



$$-2y^2 + 600y = -x^2 + 300x $$
or
$$2(y^2 - 300y) = (x^2 - 300x) $$



which produces a similar-looking graph and trivial (zero-area) results (among infinitely many): $(0,0), (0,300)$ and $(300,0)$.



enter image description here



Question from HW






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 8:24







Adam Hrankowski

















asked Dec 20 '18 at 7:33









Adam HrankowskiAdam Hrankowski

2,094930




2,094930












  • $begingroup$
    I think here the area $A$ is being treated as a known quantity. You have to select the particular $(x,y)$ that simultaneously satisfies $y^2 - 600y = x^2 - 300x,A = -x^2 + 300x,A = -y^2 + 600y$
    $endgroup$
    – Shubham Johri
    Dec 20 '18 at 7:52










  • $begingroup$
    You know what? I had -2y^2 and I "fixed" it.
    $endgroup$
    – Adam Hrankowski
    Dec 20 '18 at 8:05


















  • $begingroup$
    I think here the area $A$ is being treated as a known quantity. You have to select the particular $(x,y)$ that simultaneously satisfies $y^2 - 600y = x^2 - 300x,A = -x^2 + 300x,A = -y^2 + 600y$
    $endgroup$
    – Shubham Johri
    Dec 20 '18 at 7:52










  • $begingroup$
    You know what? I had -2y^2 and I "fixed" it.
    $endgroup$
    – Adam Hrankowski
    Dec 20 '18 at 8:05
















$begingroup$
I think here the area $A$ is being treated as a known quantity. You have to select the particular $(x,y)$ that simultaneously satisfies $y^2 - 600y = x^2 - 300x,A = -x^2 + 300x,A = -y^2 + 600y$
$endgroup$
– Shubham Johri
Dec 20 '18 at 7:52




$begingroup$
I think here the area $A$ is being treated as a known quantity. You have to select the particular $(x,y)$ that simultaneously satisfies $y^2 - 600y = x^2 - 300x,A = -x^2 + 300x,A = -y^2 + 600y$
$endgroup$
– Shubham Johri
Dec 20 '18 at 7:52












$begingroup$
You know what? I had -2y^2 and I "fixed" it.
$endgroup$
– Adam Hrankowski
Dec 20 '18 at 8:05




$begingroup$
You know what? I had -2y^2 and I "fixed" it.
$endgroup$
– Adam Hrankowski
Dec 20 '18 at 8:05










2 Answers
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oldest

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$begingroup$

The solution and that of the book doesn't make any sense because the variables $x$ and $y$ are not defined and the problem actually has four unknowns.



From the given data, using field dimensions $w,h,w',h'$, we have



$$begin{cases}A=wh,\2w+2h=600,\A=w'h',\w+2h'=600.end{cases}$$



Eliminating $h$ and $h'$, we are left with



$$begin{cases}A=300w-w^2,\A=300w'-dfrac{w'^2}2.end{cases}$$



You can eliminate $A$ and end-up with a single equation in two unknowns, which is indeterminate.






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    0












    $begingroup$

    Let the dimensions of the first field be $x,y$.



    $$x+y=300\A=xy=x(300-x)=300x-x^2$$



    Let the dimensions of the second field be $l,b$, and say we have two sides of length $b$.



    $$l+2b=600\A=lb=b(600-2b)=600b-2b^2$$



    So the systems $(C),(D)$ will provide us with the required dimensions.



    As for your query, I believe the area $A$ is being treated as a known quantity. We have to select that $(x,b)$ which satisfies $300x-x^2=A=600b-2b^2$ simultaneously.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

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      1












      $begingroup$

      The solution and that of the book doesn't make any sense because the variables $x$ and $y$ are not defined and the problem actually has four unknowns.



      From the given data, using field dimensions $w,h,w',h'$, we have



      $$begin{cases}A=wh,\2w+2h=600,\A=w'h',\w+2h'=600.end{cases}$$



      Eliminating $h$ and $h'$, we are left with



      $$begin{cases}A=300w-w^2,\A=300w'-dfrac{w'^2}2.end{cases}$$



      You can eliminate $A$ and end-up with a single equation in two unknowns, which is indeterminate.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The solution and that of the book doesn't make any sense because the variables $x$ and $y$ are not defined and the problem actually has four unknowns.



        From the given data, using field dimensions $w,h,w',h'$, we have



        $$begin{cases}A=wh,\2w+2h=600,\A=w'h',\w+2h'=600.end{cases}$$



        Eliminating $h$ and $h'$, we are left with



        $$begin{cases}A=300w-w^2,\A=300w'-dfrac{w'^2}2.end{cases}$$



        You can eliminate $A$ and end-up with a single equation in two unknowns, which is indeterminate.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The solution and that of the book doesn't make any sense because the variables $x$ and $y$ are not defined and the problem actually has four unknowns.



          From the given data, using field dimensions $w,h,w',h'$, we have



          $$begin{cases}A=wh,\2w+2h=600,\A=w'h',\w+2h'=600.end{cases}$$



          Eliminating $h$ and $h'$, we are left with



          $$begin{cases}A=300w-w^2,\A=300w'-dfrac{w'^2}2.end{cases}$$



          You can eliminate $A$ and end-up with a single equation in two unknowns, which is indeterminate.






          share|cite|improve this answer









          $endgroup$



          The solution and that of the book doesn't make any sense because the variables $x$ and $y$ are not defined and the problem actually has four unknowns.



          From the given data, using field dimensions $w,h,w',h'$, we have



          $$begin{cases}A=wh,\2w+2h=600,\A=w'h',\w+2h'=600.end{cases}$$



          Eliminating $h$ and $h'$, we are left with



          $$begin{cases}A=300w-w^2,\A=300w'-dfrac{w'^2}2.end{cases}$$



          You can eliminate $A$ and end-up with a single equation in two unknowns, which is indeterminate.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 8:38









          Yves DaoustYves Daoust

          129k675227




          129k675227























              0












              $begingroup$

              Let the dimensions of the first field be $x,y$.



              $$x+y=300\A=xy=x(300-x)=300x-x^2$$



              Let the dimensions of the second field be $l,b$, and say we have two sides of length $b$.



              $$l+2b=600\A=lb=b(600-2b)=600b-2b^2$$



              So the systems $(C),(D)$ will provide us with the required dimensions.



              As for your query, I believe the area $A$ is being treated as a known quantity. We have to select that $(x,b)$ which satisfies $300x-x^2=A=600b-2b^2$ simultaneously.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Let the dimensions of the first field be $x,y$.



                $$x+y=300\A=xy=x(300-x)=300x-x^2$$



                Let the dimensions of the second field be $l,b$, and say we have two sides of length $b$.



                $$l+2b=600\A=lb=b(600-2b)=600b-2b^2$$



                So the systems $(C),(D)$ will provide us with the required dimensions.



                As for your query, I believe the area $A$ is being treated as a known quantity. We have to select that $(x,b)$ which satisfies $300x-x^2=A=600b-2b^2$ simultaneously.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Let the dimensions of the first field be $x,y$.



                  $$x+y=300\A=xy=x(300-x)=300x-x^2$$



                  Let the dimensions of the second field be $l,b$, and say we have two sides of length $b$.



                  $$l+2b=600\A=lb=b(600-2b)=600b-2b^2$$



                  So the systems $(C),(D)$ will provide us with the required dimensions.



                  As for your query, I believe the area $A$ is being treated as a known quantity. We have to select that $(x,b)$ which satisfies $300x-x^2=A=600b-2b^2$ simultaneously.






                  share|cite|improve this answer









                  $endgroup$



                  Let the dimensions of the first field be $x,y$.



                  $$x+y=300\A=xy=x(300-x)=300x-x^2$$



                  Let the dimensions of the second field be $l,b$, and say we have two sides of length $b$.



                  $$l+2b=600\A=lb=b(600-2b)=600b-2b^2$$



                  So the systems $(C),(D)$ will provide us with the required dimensions.



                  As for your query, I believe the area $A$ is being treated as a known quantity. We have to select that $(x,b)$ which satisfies $300x-x^2=A=600b-2b^2$ simultaneously.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 20 '18 at 7:55









                  Shubham JohriShubham Johri

                  5,204718




                  5,204718






























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