How to find the largest difference in height between two lines on a graph? (one is non-linear)
$begingroup$
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
edited Jan 5 at 18:26
Ross Millikan
295k23198371
asked Jan 5 at 18:09
SpagBolSpagBol
32
$endgroup$
add a comment |
$begingroup$
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
edited Jan 5 at 18:26
Ross Millikan
295k23198371
asked Jan 5 at 18:09
SpagBolSpagBol
32
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Jan 5 at 18:26
Ross Millikan
295k23198371
asked Jan 5 at 18:09
SpagBolSpagBol
32
$endgroup$
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
edited Jan 5 at 18:26
Ross Millikan
295k23198371
asked Jan 5 at 18:09
SpagBolSpagBol
32
edited Jan 5 at 18:26
Ross Millikan
295k23198371
asked Jan 5 at 18:09
SpagBolSpagBol
32
edited Jan 5 at 18:26
Ross Millikan
295k23198371
edited Jan 5 at 18:26
Ross Millikan
295k23198371
edited Jan 5 at 18:26
Ross Millikan
295k23198371
295k23198371
asked Jan 5 at 18:09
SpagBolSpagBol
32
asked Jan 5 at 18:09
SpagBolSpagBol
32
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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?
answered Jan 5 at 18:28
Logan TollLogan Toll
834519
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 5 at 18:26
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063001%2fhow-to-find-the-largest-difference-in-height-between-two-lines-on-a-graph-one%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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?
answered Jan 5 at 18:28
Logan TollLogan Toll
834519
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
answered Jan 5 at 18:28
Logan TollLogan Toll
834519
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
answered Jan 5 at 18:28
Logan TollLogan Toll
834519
$endgroup$
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?
answered Jan 5 at 18:28
Logan TollLogan Toll
834519
answered Jan 5 at 18:28
Logan TollLogan Toll
834519
answered Jan 5 at 18:28
Logan TollLogan Toll
834519
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 5 at 18:26
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 5 at 18:26
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 5 at 18:26
Ross MillikanRoss Millikan
295k23198371
$endgroup$
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.
answered Jan 5 at 18:26
Ross MillikanRoss Millikan
295k23198371
answered Jan 5 at 18:26
Ross MillikanRoss Millikan
295k23198371
answered Jan 5 at 18:26
Ross MillikanRoss Millikan
295k23198371
answered Jan 5 at 18:26
Ross MillikanRoss Millikan
295k23198371
295k23198371
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063001%2fhow-to-find-the-largest-difference-in-height-between-two-lines-on-a-graph-one%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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