How to find the largest difference in height between two lines on a graph? (one is non-linear)












0












$begingroup$


The functions I've used as an example is $y=20x$ and $y=5x^3$. Constraints: $yleq 40$, $0leq x leq5$.



Thanks for any help...Don't know a good tag, this is...algebra...and graphs...so algebraic-geometry...?






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried? Do you know calculus?
    $endgroup$
    – Logan Toll
    Jan 5 at 18:16










  • $begingroup$
    I have an A-level in Maths but haven't done it in a while...I really don't know what to do, tried differentiating the equations...but i don't know where that leads to...
    $endgroup$
    – SpagBol
    Jan 5 at 18:20










  • $begingroup$
    I don't understand the question It sounds almost like you want to maximize $|20x-5x^3|$ over $0leq xleq 5$, but then the constraint $yleq 40$ doesn't make any sense.
    $endgroup$
    – Ben W
    Jan 5 at 18:25


















0












$begingroup$


The functions I've used as an example is $y=20x$ and $y=5x^3$. Constraints: $yleq 40$, $0leq x leq5$.



Thanks for any help...Don't know a good tag, this is...algebra...and graphs...so algebraic-geometry...?






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried? Do you know calculus?
    $endgroup$
    – Logan Toll
    Jan 5 at 18:16










  • $begingroup$
    I have an A-level in Maths but haven't done it in a while...I really don't know what to do, tried differentiating the equations...but i don't know where that leads to...
    $endgroup$
    – SpagBol
    Jan 5 at 18:20










  • $begingroup$
    I don't understand the question It sounds almost like you want to maximize $|20x-5x^3|$ over $0leq xleq 5$, but then the constraint $yleq 40$ doesn't make any sense.
    $endgroup$
    – Ben W
    Jan 5 at 18:25
















0












0








0





$begingroup$


The functions I've used as an example is $y=20x$ and $y=5x^3$. Constraints: $yleq 40$, $0leq x leq5$.



Thanks for any help...Don't know a good tag, this is...algebra...and graphs...so algebraic-geometry...?






calculus







share|cite|improve this question











$endgroup$




The functions I've used as an example is $y=20x$ and $y=5x^3$. Constraints: $yleq 40$, $0leq x leq5$.



Thanks for any help...Don't know a good tag, this is...algebra...and graphs...so algebraic-geometry...?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 18:26









Ross Millikan

295k23198371




295k23198371










asked Jan 5 at 18:09









SpagBolSpagBol

32




32












  • $begingroup$
    What have you tried? Do you know calculus?
    $endgroup$
    – Logan Toll
    Jan 5 at 18:16










  • $begingroup$
    I have an A-level in Maths but haven't done it in a while...I really don't know what to do, tried differentiating the equations...but i don't know where that leads to...
    $endgroup$
    – SpagBol
    Jan 5 at 18:20










  • $begingroup$
    I don't understand the question It sounds almost like you want to maximize $|20x-5x^3|$ over $0leq xleq 5$, but then the constraint $yleq 40$ doesn't make any sense.
    $endgroup$
    – Ben W
    Jan 5 at 18:25




















  • $begingroup$
    What have you tried? Do you know calculus?
    $endgroup$
    – Logan Toll
    Jan 5 at 18:16










  • $begingroup$
    I have an A-level in Maths but haven't done it in a while...I really don't know what to do, tried differentiating the equations...but i don't know where that leads to...
    $endgroup$
    – SpagBol
    Jan 5 at 18:20










  • $begingroup$
    I don't understand the question It sounds almost like you want to maximize $|20x-5x^3|$ over $0leq xleq 5$, but then the constraint $yleq 40$ doesn't make any sense.
    $endgroup$
    – Ben W
    Jan 5 at 18:25


















$begingroup$
What have you tried? Do you know calculus?
$endgroup$
– Logan Toll
Jan 5 at 18:16




$begingroup$
What have you tried? Do you know calculus?
$endgroup$
– Logan Toll
Jan 5 at 18:16












$begingroup$
I have an A-level in Maths but haven't done it in a while...I really don't know what to do, tried differentiating the equations...but i don't know where that leads to...
$endgroup$
– SpagBol
Jan 5 at 18:20




$begingroup$
I have an A-level in Maths but haven't done it in a while...I really don't know what to do, tried differentiating the equations...but i don't know where that leads to...
$endgroup$
– SpagBol
Jan 5 at 18:20












$begingroup$
I don't understand the question It sounds almost like you want to maximize $|20x-5x^3|$ over $0leq xleq 5$, but then the constraint $yleq 40$ doesn't make any sense.
$endgroup$
– Ben W
Jan 5 at 18:25






$begingroup$
I don't understand the question It sounds almost like you want to maximize $|20x-5x^3|$ over $0leq xleq 5$, but then the constraint $yleq 40$ doesn't make any sense.
$endgroup$
– Ben W
Jan 5 at 18:25












2 Answers
2






active

oldest

votes


















0












$begingroup$

One way you could approach it is using calculus. You've got two functions $f(x)=20x$, and $g(x)=5x^3$, we can define a new function (the height function which we will call $h$) by $h(x)=f(x)-g(x)=5x^3-20x$. This tells us the vertical distance between the two lines you're considering. Now whenever $lvert h(x)rvert$ is maximized, we know that the vertical distance between the two functions is maximized.



Now, we know from elementary calculus, that a function will maximize at critical it's critical points -- these occur when the derivative of your function is zero or we are at an end point of our interval (in this case $x=0$ or $x=5$).
$$
frac{mathrm{d}}{mathrm{d}x}h(x)=15x^2-20
$$
So when is this going to be equal to zero?
$$
15x^2-20=0 implies x = pmsqrt{frac{20}{15}}approx pm 1.1547
$$
Since we are only interested in values of $x$ that range from $0$ to $5$ we can get rid of the $-sqrt{frac{20}{15}}$ option. Now you have three values to check:



$lvert h(0)rvert$, $leftlvert hleft(sqrt{frac{20}{15}}right)rightrvert$, and $lvert h(5)rvert$.



Which one of those is largest? Does it fit within your range of allowable $y$ values? How could you restrict the original interval so that your $y$ values stay within the appropriate bounds?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, I was along the right lines...
    $endgroup$
    – SpagBol
    Jan 5 at 18:36










  • $begingroup$
    The allowable range of $x$ is only from $0$ to $2$ because $y(2)=40$ for both graphs.
    $endgroup$
    – Ross Millikan
    Jan 5 at 18:48



















1












$begingroup$

The vertical distance between them is the absolute difference of the $y$ values, so is $|20x-5x^3|$ Take the derivative of this (the absolute value bars don't matter because you just have to subtract them in the right order), set the derivative to $0$ and solve the resulting equation. That will give you a candidate point. You also have to check the end points of the interval. Note that the restriction on $y$ reduces the allowable range of $x$ values.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

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    0












    $begingroup$

    One way you could approach it is using calculus. You've got two functions $f(x)=20x$, and $g(x)=5x^3$, we can define a new function (the height function which we will call $h$) by $h(x)=f(x)-g(x)=5x^3-20x$. This tells us the vertical distance between the two lines you're considering. Now whenever $lvert h(x)rvert$ is maximized, we know that the vertical distance between the two functions is maximized.



    Now, we know from elementary calculus, that a function will maximize at critical it's critical points -- these occur when the derivative of your function is zero or we are at an end point of our interval (in this case $x=0$ or $x=5$).
    $$
    frac{mathrm{d}}{mathrm{d}x}h(x)=15x^2-20
    $$
    So when is this going to be equal to zero?
    $$
    15x^2-20=0 implies x = pmsqrt{frac{20}{15}}approx pm 1.1547
    $$
    Since we are only interested in values of $x$ that range from $0$ to $5$ we can get rid of the $-sqrt{frac{20}{15}}$ option. Now you have three values to check:



    $lvert h(0)rvert$, $leftlvert hleft(sqrt{frac{20}{15}}right)rightrvert$, and $lvert h(5)rvert$.



    Which one of those is largest? Does it fit within your range of allowable $y$ values? How could you restrict the original interval so that your $y$ values stay within the appropriate bounds?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, I was along the right lines...
      $endgroup$
      – SpagBol
      Jan 5 at 18:36










    • $begingroup$
      The allowable range of $x$ is only from $0$ to $2$ because $y(2)=40$ for both graphs.
      $endgroup$
      – Ross Millikan
      Jan 5 at 18:48
















    0












    $begingroup$

    One way you could approach it is using calculus. You've got two functions $f(x)=20x$, and $g(x)=5x^3$, we can define a new function (the height function which we will call $h$) by $h(x)=f(x)-g(x)=5x^3-20x$. This tells us the vertical distance between the two lines you're considering. Now whenever $lvert h(x)rvert$ is maximized, we know that the vertical distance between the two functions is maximized.



    Now, we know from elementary calculus, that a function will maximize at critical it's critical points -- these occur when the derivative of your function is zero or we are at an end point of our interval (in this case $x=0$ or $x=5$).
    $$
    frac{mathrm{d}}{mathrm{d}x}h(x)=15x^2-20
    $$
    So when is this going to be equal to zero?
    $$
    15x^2-20=0 implies x = pmsqrt{frac{20}{15}}approx pm 1.1547
    $$
    Since we are only interested in values of $x$ that range from $0$ to $5$ we can get rid of the $-sqrt{frac{20}{15}}$ option. Now you have three values to check:



    $lvert h(0)rvert$, $leftlvert hleft(sqrt{frac{20}{15}}right)rightrvert$, and $lvert h(5)rvert$.



    Which one of those is largest? Does it fit within your range of allowable $y$ values? How could you restrict the original interval so that your $y$ values stay within the appropriate bounds?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, I was along the right lines...
      $endgroup$
      – SpagBol
      Jan 5 at 18:36










    • $begingroup$
      The allowable range of $x$ is only from $0$ to $2$ because $y(2)=40$ for both graphs.
      $endgroup$
      – Ross Millikan
      Jan 5 at 18:48














    0












    0








    0





    $begingroup$

    One way you could approach it is using calculus. You've got two functions $f(x)=20x$, and $g(x)=5x^3$, we can define a new function (the height function which we will call $h$) by $h(x)=f(x)-g(x)=5x^3-20x$. This tells us the vertical distance between the two lines you're considering. Now whenever $lvert h(x)rvert$ is maximized, we know that the vertical distance between the two functions is maximized.



    Now, we know from elementary calculus, that a function will maximize at critical it's critical points -- these occur when the derivative of your function is zero or we are at an end point of our interval (in this case $x=0$ or $x=5$).
    $$
    frac{mathrm{d}}{mathrm{d}x}h(x)=15x^2-20
    $$
    So when is this going to be equal to zero?
    $$
    15x^2-20=0 implies x = pmsqrt{frac{20}{15}}approx pm 1.1547
    $$
    Since we are only interested in values of $x$ that range from $0$ to $5$ we can get rid of the $-sqrt{frac{20}{15}}$ option. Now you have three values to check:



    $lvert h(0)rvert$, $leftlvert hleft(sqrt{frac{20}{15}}right)rightrvert$, and $lvert h(5)rvert$.



    Which one of those is largest? Does it fit within your range of allowable $y$ values? How could you restrict the original interval so that your $y$ values stay within the appropriate bounds?






    share|cite|improve this answer









    $endgroup$



    One way you could approach it is using calculus. You've got two functions $f(x)=20x$, and $g(x)=5x^3$, we can define a new function (the height function which we will call $h$) by $h(x)=f(x)-g(x)=5x^3-20x$. This tells us the vertical distance between the two lines you're considering. Now whenever $lvert h(x)rvert$ is maximized, we know that the vertical distance between the two functions is maximized.



    Now, we know from elementary calculus, that a function will maximize at critical it's critical points -- these occur when the derivative of your function is zero or we are at an end point of our interval (in this case $x=0$ or $x=5$).
    $$
    frac{mathrm{d}}{mathrm{d}x}h(x)=15x^2-20
    $$
    So when is this going to be equal to zero?
    $$
    15x^2-20=0 implies x = pmsqrt{frac{20}{15}}approx pm 1.1547
    $$
    Since we are only interested in values of $x$ that range from $0$ to $5$ we can get rid of the $-sqrt{frac{20}{15}}$ option. Now you have three values to check:



    $lvert h(0)rvert$, $leftlvert hleft(sqrt{frac{20}{15}}right)rightrvert$, and $lvert h(5)rvert$.



    Which one of those is largest? Does it fit within your range of allowable $y$ values? How could you restrict the original interval so that your $y$ values stay within the appropriate bounds?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 5 at 18:28









    Logan TollLogan Toll

    834519




    834519












    • $begingroup$
      Thank you, I was along the right lines...
      $endgroup$
      – SpagBol
      Jan 5 at 18:36










    • $begingroup$
      The allowable range of $x$ is only from $0$ to $2$ because $y(2)=40$ for both graphs.
      $endgroup$
      – Ross Millikan
      Jan 5 at 18:48


















    • $begingroup$
      Thank you, I was along the right lines...
      $endgroup$
      – SpagBol
      Jan 5 at 18:36










    • $begingroup$
      The allowable range of $x$ is only from $0$ to $2$ because $y(2)=40$ for both graphs.
      $endgroup$
      – Ross Millikan
      Jan 5 at 18:48
















    $begingroup$
    Thank you, I was along the right lines...
    $endgroup$
    – SpagBol
    Jan 5 at 18:36




    $begingroup$
    Thank you, I was along the right lines...
    $endgroup$
    – SpagBol
    Jan 5 at 18:36












    $begingroup$
    The allowable range of $x$ is only from $0$ to $2$ because $y(2)=40$ for both graphs.
    $endgroup$
    – Ross Millikan
    Jan 5 at 18:48




    $begingroup$
    The allowable range of $x$ is only from $0$ to $2$ because $y(2)=40$ for both graphs.
    $endgroup$
    – Ross Millikan
    Jan 5 at 18:48











    1












    $begingroup$

    The vertical distance between them is the absolute difference of the $y$ values, so is $|20x-5x^3|$ Take the derivative of this (the absolute value bars don't matter because you just have to subtract them in the right order), set the derivative to $0$ and solve the resulting equation. That will give you a candidate point. You also have to check the end points of the interval. Note that the restriction on $y$ reduces the allowable range of $x$ values.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The vertical distance between them is the absolute difference of the $y$ values, so is $|20x-5x^3|$ Take the derivative of this (the absolute value bars don't matter because you just have to subtract them in the right order), set the derivative to $0$ and solve the resulting equation. That will give you a candidate point. You also have to check the end points of the interval. Note that the restriction on $y$ reduces the allowable range of $x$ values.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The vertical distance between them is the absolute difference of the $y$ values, so is $|20x-5x^3|$ Take the derivative of this (the absolute value bars don't matter because you just have to subtract them in the right order), set the derivative to $0$ and solve the resulting equation. That will give you a candidate point. You also have to check the end points of the interval. Note that the restriction on $y$ reduces the allowable range of $x$ values.






        share|cite|improve this answer









        $endgroup$



        The vertical distance between them is the absolute difference of the $y$ values, so is $|20x-5x^3|$ Take the derivative of this (the absolute value bars don't matter because you just have to subtract them in the right order), set the derivative to $0$ and solve the resulting equation. That will give you a candidate point. You also have to check the end points of the interval. Note that the restriction on $y$ reduces the allowable range of $x$ values.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 18:26









        Ross MillikanRoss Millikan

        295k23198371




        295k23198371






























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