Sensitivity of revenue to price
$begingroup$
This is a rather simple problem, but i can't for the life of me figure out the logic behind it.
The revenue $R $ from a software product depends on the price $p$ charged by the distributor according to the formula.
$$R = 4000p-10p^2$$
How sensitive is $R$ to $p$ when p is a) $100 $, b)$200$ c)$300 $?
Which begs for the differentiation: $frac {dR}{dp}=4000-20p$
Now, finding the rate of change is easy. The rate of change is $2000$, $0$, $-2000$ for the respective values $100$, $200$, $300$.
If we were to maximize the revenue we would go for the $200$ option, because it is an absolute maximum value in the function $R = 4000p-10p^2$.
But practically, doesn't it mean that the revenue is $0$ dollars per dollar charged. And that a price at $200$ dollars would give us the revenue of $2000$ dollars per dollar charged?
derivatives
asked Jan 1 at 18:04
BullerskyddBullerskydd
43
$endgroup$
add a comment |
$begingroup$
This is a rather simple problem, but i can't for the life of me figure out the logic behind it.
The revenue $R $ from a software product depends on the price $p$ charged by the distributor according to the formula.
$$R = 4000p-10p^2$$
How sensitive is $R$ to $p$ when p is a) $100 $, b)$200$ c)$300 $?
Which begs for the differentiation: $frac {dR}{dp}=4000-20p$
Now, finding the rate of change is easy. The rate of change is $2000$, $0$, $-2000$ for the respective values $100$, $200$, $300$.
If we were to maximize the revenue we would go for the $200$ option, because it is an absolute maximum value in the function $R = 4000p-10p^2$.
But practically, doesn't it mean that the revenue is $0$ dollars per dollar charged. And that a price at $200$ dollars would give us the revenue of $2000$ dollars per dollar charged?
derivatives
asked Jan 1 at 18:04
BullerskyddBullerskydd
43
$endgroup$
$begingroup$
Note: I reformatted your post. The version of Latex used by this site uses "$$'s $" to indicate code intended for compiling, not as dollar signs. You can insert dollar signs, as I did in this comment, but it's usually more trouble than it is worth.
$endgroup$
– lulu
Jan 1 at 18:09
$begingroup$
That said, I don't understand your question. The maximum revenue comes when the derivative vanishes (as there is clearly a global max and no global min). Thus at $p=200$, as you point out. But at $p=200$ we have $R(200)=4000times 200 -10times 200^2=400,000$ Where is the confusion?
$endgroup$
– lulu
Jan 1 at 18:12
$begingroup$
The concept of rate of change is what set me up for confusion. I suppose it is synonymous with the slope at a given point. The slope at p = 100 is 2000. Doesn't this mean that for every dollar charged at this point, the income is 2000 dollars? Where as for p=200, where the slope is 0, there is 0 income for every dollar spent. Now I get that the maximum revenue is when the function reaches its peak. But how would we go about finding maximum revenue if there was no absolute maximum value at dR/dp=0? What if it would have been an inflection point?
$endgroup$
– Bullerskydd
Jan 1 at 18:25
$begingroup$
Well, it's certainly true that this case is much simpler than the general case. Calculus isn't even needed...just complete the squares to write $R(p)=-10(p-200)^2+400000$ to make the global max visible. In general one does need to look at a second derivative, or nearby values, to test critical points. Not necessary here.
$endgroup$
– lulu
Jan 1 at 18:27
add a comment |
$begingroup$
This is a rather simple problem, but i can't for the life of me figure out the logic behind it.
The revenue $R $ from a software product depends on the price $p$ charged by the distributor according to the formula.
$$R = 4000p-10p^2$$
How sensitive is $R$ to $p$ when p is a) $100 $, b)$200$ c)$300 $?
Which begs for the differentiation: $frac {dR}{dp}=4000-20p$
Now, finding the rate of change is easy. The rate of change is $2000$, $0$, $-2000$ for the respective values $100$, $200$, $300$.
If we were to maximize the revenue we would go for the $200$ option, because it is an absolute maximum value in the function $R = 4000p-10p^2$.
But practically, doesn't it mean that the revenue is $0$ dollars per dollar charged. And that a price at $200$ dollars would give us the revenue of $2000$ dollars per dollar charged?
derivatives
asked Jan 1 at 18:04
BullerskyddBullerskydd
43
$endgroup$
This is a rather simple problem, but i can't for the life of me figure out the logic behind it.
The revenue $R $ from a software product depends on the price $p$ charged by the distributor according to the formula.
$$R = 4000p-10p^2$$
How sensitive is $R$ to $p$ when p is a) $100 $, b)$200$ c)$300 $?
Which begs for the differentiation: $frac {dR}{dp}=4000-20p$
Now, finding the rate of change is easy. The rate of change is $2000$, $0$, $-2000$ for the respective values $100$, $200$, $300$.
If we were to maximize the revenue we would go for the $200$ option, because it is an absolute maximum value in the function $R = 4000p-10p^2$.
But practically, doesn't it mean that the revenue is $0$ dollars per dollar charged. And that a price at $200$ dollars would give us the revenue of $2000$ dollars per dollar charged?
derivatives
derivatives
asked Jan 1 at 18:04
BullerskyddBullerskydd
43
asked Jan 1 at 18:04
BullerskyddBullerskydd
43
edited Jan 1 at 18:12
Bullerskydd
asked Jan 1 at 18:04
BullerskyddBullerskydd
43
asked Jan 1 at 18:04
BullerskyddBullerskydd
43
asked Jan 1 at 18:04
BullerskyddBullerskydd
43
43
$begingroup$
Note: I reformatted your post. The version of Latex used by this site uses "$$'s $" to indicate code intended for compiling, not as dollar signs. You can insert dollar signs, as I did in this comment, but it's usually more trouble than it is worth.
$endgroup$
– lulu
Jan 1 at 18:09
$begingroup$
That said, I don't understand your question. The maximum revenue comes when the derivative vanishes (as there is clearly a global max and no global min). Thus at $p=200$, as you point out. But at $p=200$ we have $R(200)=4000times 200 -10times 200^2=400,000$ Where is the confusion?
$endgroup$
– lulu
Jan 1 at 18:12
$begingroup$
The concept of rate of change is what set me up for confusion. I suppose it is synonymous with the slope at a given point. The slope at p = 100 is 2000. Doesn't this mean that for every dollar charged at this point, the income is 2000 dollars? Where as for p=200, where the slope is 0, there is 0 income for every dollar spent. Now I get that the maximum revenue is when the function reaches its peak. But how would we go about finding maximum revenue if there was no absolute maximum value at dR/dp=0? What if it would have been an inflection point?
$endgroup$
– Bullerskydd
Jan 1 at 18:25
$begingroup$
Well, it's certainly true that this case is much simpler than the general case. Calculus isn't even needed...just complete the squares to write $R(p)=-10(p-200)^2+400000$ to make the global max visible. In general one does need to look at a second derivative, or nearby values, to test critical points. Not necessary here.
$endgroup$
– lulu
Jan 1 at 18:27
add a comment |
$begingroup$
Note: I reformatted your post. The version of Latex used by this site uses "$$'s $" to indicate code intended for compiling, not as dollar signs. You can insert dollar signs, as I did in this comment, but it's usually more trouble than it is worth.
$endgroup$
– lulu
Jan 1 at 18:09
$begingroup$
That said, I don't understand your question. The maximum revenue comes when the derivative vanishes (as there is clearly a global max and no global min). Thus at $p=200$, as you point out. But at $p=200$ we have $R(200)=4000times 200 -10times 200^2=400,000$ Where is the confusion?
$endgroup$
– lulu
Jan 1 at 18:12
$begingroup$
The concept of rate of change is what set me up for confusion. I suppose it is synonymous with the slope at a given point. The slope at p = 100 is 2000. Doesn't this mean that for every dollar charged at this point, the income is 2000 dollars? Where as for p=200, where the slope is 0, there is 0 income for every dollar spent. Now I get that the maximum revenue is when the function reaches its peak. But how would we go about finding maximum revenue if there was no absolute maximum value at dR/dp=0? What if it would have been an inflection point?
$endgroup$
– Bullerskydd
Jan 1 at 18:25
$begingroup$
Well, it's certainly true that this case is much simpler than the general case. Calculus isn't even needed...just complete the squares to write $R(p)=-10(p-200)^2+400000$ to make the global max visible. In general one does need to look at a second derivative, or nearby values, to test critical points. Not necessary here.
$endgroup$
– lulu
Jan 1 at 18:27
$begingroup$
Note: I reformatted your post. The version of Latex used by this site uses "$$'s $" to indicate code intended for compiling, not as dollar signs. You can insert dollar signs, as I did in this comment, but it's usually more trouble than it is worth.
$endgroup$
– lulu
Jan 1 at 18:09
$begingroup$
Note: I reformatted your post. The version of Latex used by this site uses "$$'s $" to indicate code intended for compiling, not as dollar signs. You can insert dollar signs, as I did in this comment, but it's usually more trouble than it is worth.
$endgroup$
– lulu
Jan 1 at 18:09
$begingroup$
That said, I don't understand your question. The maximum revenue comes when the derivative vanishes (as there is clearly a global max and no global min). Thus at $p=200$, as you point out. But at $p=200$ we have $R(200)=4000times 200 -10times 200^2=400,000$ Where is the confusion?
$endgroup$
– lulu
Jan 1 at 18:12
$begingroup$
That said, I don't understand your question. The maximum revenue comes when the derivative vanishes (as there is clearly a global max and no global min). Thus at $p=200$, as you point out. But at $p=200$ we have $R(200)=4000times 200 -10times 200^2=400,000$ Where is the confusion?
$endgroup$
– lulu
Jan 1 at 18:12
$begingroup$
The concept of rate of change is what set me up for confusion. I suppose it is synonymous with the slope at a given point. The slope at p = 100 is 2000. Doesn't this mean that for every dollar charged at this point, the income is 2000 dollars? Where as for p=200, where the slope is 0, there is 0 income for every dollar spent. Now I get that the maximum revenue is when the function reaches its peak. But how would we go about finding maximum revenue if there was no absolute maximum value at dR/dp=0? What if it would have been an inflection point?
$endgroup$
– Bullerskydd
Jan 1 at 18:25
$begingroup$
The concept of rate of change is what set me up for confusion. I suppose it is synonymous with the slope at a given point. The slope at p = 100 is 2000. Doesn't this mean that for every dollar charged at this point, the income is 2000 dollars? Where as for p=200, where the slope is 0, there is 0 income for every dollar spent. Now I get that the maximum revenue is when the function reaches its peak. But how would we go about finding maximum revenue if there was no absolute maximum value at dR/dp=0? What if it would have been an inflection point?
$endgroup$
– Bullerskydd
Jan 1 at 18:25
$begingroup$
Well, it's certainly true that this case is much simpler than the general case. Calculus isn't even needed...just complete the squares to write $R(p)=-10(p-200)^2+400000$ to make the global max visible. In general one does need to look at a second derivative, or nearby values, to test critical points. Not necessary here.
$endgroup$
– lulu
Jan 1 at 18:27
$begingroup$
Well, it's certainly true that this case is much simpler than the general case. Calculus isn't even needed...just complete the squares to write $R(p)=-10(p-200)^2+400000$ to make the global max visible. In general one does need to look at a second derivative, or nearby values, to test critical points. Not necessary here.
$endgroup$
– lulu
Jan 1 at 18:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Plot the revenue curve and see how revenue changes at random $dp$ intervals on the price axis. At the maxima it barely changes with small price shifts, right? Since growth rate of parabola is very small at the start. Now check some remote intervals
answered Jan 1 at 18:20
MakinaMakina
1,1581316
$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%2f3058709%2fsensitivity-of-revenue-to-price%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Plot the revenue curve and see how revenue changes at random $dp$ intervals on the price axis. At the maxima it barely changes with small price shifts, right? Since growth rate of parabola is very small at the start. Now check some remote intervals
answered Jan 1 at 18:20
MakinaMakina
1,1581316
$endgroup$
add a comment |
$begingroup$
Plot the revenue curve and see how revenue changes at random $dp$ intervals on the price axis. At the maxima it barely changes with small price shifts, right? Since growth rate of parabola is very small at the start. Now check some remote intervals
answered Jan 1 at 18:20
MakinaMakina
1,1581316
$endgroup$
add a comment |
$begingroup$
Plot the revenue curve and see how revenue changes at random $dp$ intervals on the price axis. At the maxima it barely changes with small price shifts, right? Since growth rate of parabola is very small at the start. Now check some remote intervals
answered Jan 1 at 18:20
MakinaMakina
1,1581316
$endgroup$
Plot the revenue curve and see how revenue changes at random $dp$ intervals on the price axis. At the maxima it barely changes with small price shifts, right? Since growth rate of parabola is very small at the start. Now check some remote intervals
answered Jan 1 at 18:20
MakinaMakina
1,1581316
answered Jan 1 at 18:20
MakinaMakina
1,1581316
answered Jan 1 at 18:20
MakinaMakina
1,1581316
answered Jan 1 at 18:20
MakinaMakina
1,1581316
1,1581316
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%2f3058709%2fsensitivity-of-revenue-to-price%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$
Note: I reformatted your post. The version of Latex used by this site uses "$$'s $" to indicate code intended for compiling, not as dollar signs. You can insert dollar signs, as I did in this comment, but it's usually more trouble than it is worth.
$endgroup$
– lulu
Jan 1 at 18:09
$begingroup$
That said, I don't understand your question. The maximum revenue comes when the derivative vanishes (as there is clearly a global max and no global min). Thus at $p=200$, as you point out. But at $p=200$ we have $R(200)=4000times 200 -10times 200^2=400,000$ Where is the confusion?
$endgroup$
– lulu
Jan 1 at 18:12
$begingroup$
The concept of rate of change is what set me up for confusion. I suppose it is synonymous with the slope at a given point. The slope at p = 100 is 2000. Doesn't this mean that for every dollar charged at this point, the income is 2000 dollars? Where as for p=200, where the slope is 0, there is 0 income for every dollar spent. Now I get that the maximum revenue is when the function reaches its peak. But how would we go about finding maximum revenue if there was no absolute maximum value at dR/dp=0? What if it would have been an inflection point?
$endgroup$
– Bullerskydd
Jan 1 at 18:25
$begingroup$
Well, it's certainly true that this case is much simpler than the general case. Calculus isn't even needed...just complete the squares to write $R(p)=-10(p-200)^2+400000$ to make the global max visible. In general one does need to look at a second derivative, or nearby values, to test critical points. Not necessary here.
$endgroup$
– lulu
Jan 1 at 18:27