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












0














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







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  • What have you tried? Do you know calculus?
    – Logan Toll
    2 days ago










  • 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...
    – SpagBol
    2 days ago










  • 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.
    – Ben W
    2 days ago


















0














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









New contributor




SpagBol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • What have you tried? Do you know calculus?
    – Logan Toll
    2 days ago










  • 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...
    – SpagBol
    2 days ago










  • 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.
    – Ben W
    2 days ago
















0












0








0







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









New contributor




SpagBol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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









New contributor




SpagBol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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SpagBol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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edited 2 days ago









Ross Millikan

292k23197371




292k23197371






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asked 2 days ago









SpagBolSpagBol

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SpagBol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.












  • What have you tried? Do you know calculus?
    – Logan Toll
    2 days ago










  • 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...
    – SpagBol
    2 days ago










  • 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.
    – Ben W
    2 days ago




















  • What have you tried? Do you know calculus?
    – Logan Toll
    2 days ago










  • 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...
    – SpagBol
    2 days ago










  • 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.
    – Ben W
    2 days ago


















What have you tried? Do you know calculus?
– Logan Toll
2 days ago




What have you tried? Do you know calculus?
– Logan Toll
2 days ago












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...
– SpagBol
2 days ago




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...
– SpagBol
2 days ago












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.
– Ben W
2 days ago






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.
– Ben W
2 days ago












2 Answers
2






active

oldest

votes


















0














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





















  • Thank you, I was along the right lines...
    – SpagBol
    2 days ago










  • The allowable range of $x$ is only from $0$ to $2$ because $y(2)=40$ for both graphs.
    – Ross Millikan
    2 days ago



















1














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
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    active

    oldest

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    2 Answers
    2






    active

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    active

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    active

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    0














    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





















    • Thank you, I was along the right lines...
      – SpagBol
      2 days ago










    • The allowable range of $x$ is only from $0$ to $2$ because $y(2)=40$ for both graphs.
      – Ross Millikan
      2 days ago
















    0














    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





















    • Thank you, I was along the right lines...
      – SpagBol
      2 days ago










    • The allowable range of $x$ is only from $0$ to $2$ because $y(2)=40$ for both graphs.
      – Ross Millikan
      2 days ago














    0












    0








    0






    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












    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 2 days ago









    Logan TollLogan Toll

    834518




    834518












    • Thank you, I was along the right lines...
      – SpagBol
      2 days ago










    • The allowable range of $x$ is only from $0$ to $2$ because $y(2)=40$ for both graphs.
      – Ross Millikan
      2 days ago


















    • Thank you, I was along the right lines...
      – SpagBol
      2 days ago










    • The allowable range of $x$ is only from $0$ to $2$ because $y(2)=40$ for both graphs.
      – Ross Millikan
      2 days ago
















    Thank you, I was along the right lines...
    – SpagBol
    2 days ago




    Thank you, I was along the right lines...
    – SpagBol
    2 days ago












    The allowable range of $x$ is only from $0$ to $2$ because $y(2)=40$ for both graphs.
    – Ross Millikan
    2 days ago




    The allowable range of $x$ is only from $0$ to $2$ because $y(2)=40$ for both graphs.
    – Ross Millikan
    2 days ago











    1














    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


























      1














      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
























        1












        1








        1






        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












        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 2 days ago









        Ross MillikanRoss Millikan

        292k23197371




        292k23197371






















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