Maximizing profit function given cost and demand functions
$begingroup$
I am given the demand function $$D(x)=10x^2 + 50x$$ and a total cost of $$C(x) = x^3 + 10x$$ where $x$ is the number of units demanded. I am asked to maximize the profit so what I did is I used the formula $$P(x) = R(x) - C(x)$$ where $R(x) = x D(x)$. Then I took the derivative of $P(x)$ and equated to 0. But I got a negative value of x more so a value less than 1 since $$P'(x) = 27x^2 + 100x -10$$ Am I right or did I do something wrong in between my process? I don't think it's logical to have a quantity which is negative and less than 1 for this kind of problem. Please help.
calculus derivatives proof-verification optimization economics
edited Dec 18 '18 at 9:43
Batominovski
33.1k33293
asked May 24 '15 at 8:54
marg_ocruzmarg_ocruz
301211
$endgroup$
add a comment |
$begingroup$
I am given the demand function $$D(x)=10x^2 + 50x$$ and a total cost of $$C(x) = x^3 + 10x$$ where $x$ is the number of units demanded. I am asked to maximize the profit so what I did is I used the formula $$P(x) = R(x) - C(x)$$ where $R(x) = x D(x)$. Then I took the derivative of $P(x)$ and equated to 0. But I got a negative value of x more so a value less than 1 since $$P'(x) = 27x^2 + 100x -10$$ Am I right or did I do something wrong in between my process? I don't think it's logical to have a quantity which is negative and less than 1 for this kind of problem. Please help.
calculus derivatives proof-verification optimization economics
edited Dec 18 '18 at 9:43
Batominovski
33.1k33293
asked May 24 '15 at 8:54
marg_ocruzmarg_ocruz
301211
$endgroup$
add a comment |
$begingroup$
I am given the demand function $$D(x)=10x^2 + 50x$$ and a total cost of $$C(x) = x^3 + 10x$$ where $x$ is the number of units demanded. I am asked to maximize the profit so what I did is I used the formula $$P(x) = R(x) - C(x)$$ where $R(x) = x D(x)$. Then I took the derivative of $P(x)$ and equated to 0. But I got a negative value of x more so a value less than 1 since $$P'(x) = 27x^2 + 100x -10$$ Am I right or did I do something wrong in between my process? I don't think it's logical to have a quantity which is negative and less than 1 for this kind of problem. Please help.
calculus derivatives proof-verification optimization economics
edited Dec 18 '18 at 9:43
Batominovski
33.1k33293
asked May 24 '15 at 8:54
marg_ocruzmarg_ocruz
301211
$endgroup$
I am given the demand function $$D(x)=10x^2 + 50x$$ and a total cost of $$C(x) = x^3 + 10x$$ where $x$ is the number of units demanded. I am asked to maximize the profit so what I did is I used the formula $$P(x) = R(x) - C(x)$$ where $R(x) = x D(x)$. Then I took the derivative of $P(x)$ and equated to 0. But I got a negative value of x more so a value less than 1 since $$P'(x) = 27x^2 + 100x -10$$ Am I right or did I do something wrong in between my process? I don't think it's logical to have a quantity which is negative and less than 1 for this kind of problem. Please help.
calculus derivatives proof-verification optimization economics
calculus derivatives proof-verification optimization economics
edited Dec 18 '18 at 9:43
Batominovski
33.1k33293
asked May 24 '15 at 8:54
marg_ocruzmarg_ocruz
301211
edited Dec 18 '18 at 9:43
Batominovski
33.1k33293
asked May 24 '15 at 8:54
marg_ocruzmarg_ocruz
301211
edited Dec 18 '18 at 9:43
Batominovski
33.1k33293
edited Dec 18 '18 at 9:43
Batominovski
33.1k33293
edited Dec 18 '18 at 9:43
Batominovski
33.1k33293
33.1k33293
asked May 24 '15 at 8:54
marg_ocruzmarg_ocruz
301211
asked May 24 '15 at 8:54
marg_ocruzmarg_ocruz
301211
asked May 24 '15 at 8:54
marg_ocruzmarg_ocruz
301211
301211
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
You should multiply $D(x)$ by the price of a single unit, to get the total revenue.
So if $Q$ is the price of a single unit (which presumably does not depend on $x$), you get
$$P(x) = QD(x) - C(x)$$
After that you are correct,to maximize this value with respect to $x$ you take the derivative and equate it to $0$ (you should show that this is indeed a maximum, and not a minimum or a stationary point. But it's not difficult to do so)
answered May 24 '15 at 9:01
AntAnt
17.4k22874
$endgroup$
$begingroup$
how do i get $Q$? I am not given the value of $Q$... :(
$endgroup$
– marg_ocruz
May 24 '15 at 9:13
$begingroup$
@marg_ocruz You should probably explain your question better.. Because you wrote that $x$ is the "number of units demanded" and that $D(x)$ is the "demand function".. What is the difference? It's not clear to me :-)
$endgroup$
– Ant
May 24 '15 at 9:15
$begingroup$
@marg_ocruz If instead $x$ is the price of a unit (which makes sense, as both $D$ is going to depend on $x$) then your calculations are correct. In this case that value you get implies setting the price to something much less than $1$.. Irrealistic, but this is just a paper problem :-)
$endgroup$
– Ant
May 24 '15 at 9:16
$begingroup$
the problem states that x is the number of units demanded and that D(x) is the demand function. I think it's mistakenly written by my professor.. I got your point and thanks a lot.. :)
$endgroup$
– marg_ocruz
May 24 '15 at 9:21
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You should multiply $D(x)$ by the price of a single unit, to get the total revenue.
So if $Q$ is the price of a single unit (which presumably does not depend on $x$), you get
$$P(x) = QD(x) - C(x)$$
After that you are correct,to maximize this value with respect to $x$ you take the derivative and equate it to $0$ (you should show that this is indeed a maximum, and not a minimum or a stationary point. But it's not difficult to do so)
answered May 24 '15 at 9:01
AntAnt
17.4k22874
$endgroup$
$begingroup$
how do i get $Q$? I am not given the value of $Q$... :(
$endgroup$
– marg_ocruz
May 24 '15 at 9:13
$begingroup$
@marg_ocruz You should probably explain your question better.. Because you wrote that $x$ is the "number of units demanded" and that $D(x)$ is the "demand function".. What is the difference? It's not clear to me :-)
$endgroup$
– Ant
May 24 '15 at 9:15
$begingroup$
@marg_ocruz If instead $x$ is the price of a unit (which makes sense, as both $D$ is going to depend on $x$) then your calculations are correct. In this case that value you get implies setting the price to something much less than $1$.. Irrealistic, but this is just a paper problem :-)
$endgroup$
– Ant
May 24 '15 at 9:16
$begingroup$
the problem states that x is the number of units demanded and that D(x) is the demand function. I think it's mistakenly written by my professor.. I got your point and thanks a lot.. :)
$endgroup$
– marg_ocruz
May 24 '15 at 9:21
add a comment |
$begingroup$
You should multiply $D(x)$ by the price of a single unit, to get the total revenue.
So if $Q$ is the price of a single unit (which presumably does not depend on $x$), you get
$$P(x) = QD(x) - C(x)$$
After that you are correct,to maximize this value with respect to $x$ you take the derivative and equate it to $0$ (you should show that this is indeed a maximum, and not a minimum or a stationary point. But it's not difficult to do so)
answered May 24 '15 at 9:01
AntAnt
17.4k22874
$endgroup$
$begingroup$
how do i get $Q$? I am not given the value of $Q$... :(
$endgroup$
– marg_ocruz
May 24 '15 at 9:13
$begingroup$
@marg_ocruz You should probably explain your question better.. Because you wrote that $x$ is the "number of units demanded" and that $D(x)$ is the "demand function".. What is the difference? It's not clear to me :-)
$endgroup$
– Ant
May 24 '15 at 9:15
$begingroup$
@marg_ocruz If instead $x$ is the price of a unit (which makes sense, as both $D$ is going to depend on $x$) then your calculations are correct. In this case that value you get implies setting the price to something much less than $1$.. Irrealistic, but this is just a paper problem :-)
$endgroup$
– Ant
May 24 '15 at 9:16
$begingroup$
the problem states that x is the number of units demanded and that D(x) is the demand function. I think it's mistakenly written by my professor.. I got your point and thanks a lot.. :)
$endgroup$
– marg_ocruz
May 24 '15 at 9:21
add a comment |
$begingroup$
You should multiply $D(x)$ by the price of a single unit, to get the total revenue.
So if $Q$ is the price of a single unit (which presumably does not depend on $x$), you get
$$P(x) = QD(x) - C(x)$$
After that you are correct,to maximize this value with respect to $x$ you take the derivative and equate it to $0$ (you should show that this is indeed a maximum, and not a minimum or a stationary point. But it's not difficult to do so)
answered May 24 '15 at 9:01
AntAnt
17.4k22874
$endgroup$
You should multiply $D(x)$ by the price of a single unit, to get the total revenue.
So if $Q$ is the price of a single unit (which presumably does not depend on $x$), you get
$$P(x) = QD(x) - C(x)$$
After that you are correct,to maximize this value with respect to $x$ you take the derivative and equate it to $0$ (you should show that this is indeed a maximum, and not a minimum or a stationary point. But it's not difficult to do so)
answered May 24 '15 at 9:01
AntAnt
17.4k22874
answered May 24 '15 at 9:01
AntAnt
17.4k22874
answered May 24 '15 at 9:01
AntAnt
17.4k22874
answered May 24 '15 at 9:01
AntAnt
17.4k22874
17.4k22874
$begingroup$
how do i get $Q$? I am not given the value of $Q$... :(
$endgroup$
– marg_ocruz
May 24 '15 at 9:13
$begingroup$
@marg_ocruz You should probably explain your question better.. Because you wrote that $x$ is the "number of units demanded" and that $D(x)$ is the "demand function".. What is the difference? It's not clear to me :-)
$endgroup$
– Ant
May 24 '15 at 9:15
$begingroup$
@marg_ocruz If instead $x$ is the price of a unit (which makes sense, as both $D$ is going to depend on $x$) then your calculations are correct. In this case that value you get implies setting the price to something much less than $1$.. Irrealistic, but this is just a paper problem :-)
$endgroup$
– Ant
May 24 '15 at 9:16
$begingroup$
the problem states that x is the number of units demanded and that D(x) is the demand function. I think it's mistakenly written by my professor.. I got your point and thanks a lot.. :)
$endgroup$
– marg_ocruz
May 24 '15 at 9:21
add a comment |
$begingroup$
how do i get $Q$? I am not given the value of $Q$... :(
$endgroup$
– marg_ocruz
May 24 '15 at 9:13
$begingroup$
@marg_ocruz You should probably explain your question better.. Because you wrote that $x$ is the "number of units demanded" and that $D(x)$ is the "demand function".. What is the difference? It's not clear to me :-)
$endgroup$
– Ant
May 24 '15 at 9:15
$begingroup$
@marg_ocruz If instead $x$ is the price of a unit (which makes sense, as both $D$ is going to depend on $x$) then your calculations are correct. In this case that value you get implies setting the price to something much less than $1$.. Irrealistic, but this is just a paper problem :-)
$endgroup$
– Ant
May 24 '15 at 9:16
$begingroup$
the problem states that x is the number of units demanded and that D(x) is the demand function. I think it's mistakenly written by my professor.. I got your point and thanks a lot.. :)
$endgroup$
– marg_ocruz
May 24 '15 at 9:21
$begingroup$
how do i get $Q$? I am not given the value of $Q$... :(
$endgroup$
– marg_ocruz
May 24 '15 at 9:13
$begingroup$
how do i get $Q$? I am not given the value of $Q$... :(
$endgroup$
– marg_ocruz
May 24 '15 at 9:13
$begingroup$
@marg_ocruz You should probably explain your question better.. Because you wrote that $x$ is the "number of units demanded" and that $D(x)$ is the "demand function".. What is the difference? It's not clear to me :-)
$endgroup$
– Ant
May 24 '15 at 9:15
$begingroup$
@marg_ocruz You should probably explain your question better.. Because you wrote that $x$ is the "number of units demanded" and that $D(x)$ is the "demand function".. What is the difference? It's not clear to me :-)
$endgroup$
– Ant
May 24 '15 at 9:15
$begingroup$
@marg_ocruz If instead $x$ is the price of a unit (which makes sense, as both $D$ is going to depend on $x$) then your calculations are correct. In this case that value you get implies setting the price to something much less than $1$.. Irrealistic, but this is just a paper problem :-)
$endgroup$
– Ant
May 24 '15 at 9:16
$begingroup$
@marg_ocruz If instead $x$ is the price of a unit (which makes sense, as both $D$ is going to depend on $x$) then your calculations are correct. In this case that value you get implies setting the price to something much less than $1$.. Irrealistic, but this is just a paper problem :-)
$endgroup$
– Ant
May 24 '15 at 9:16
$begingroup$
the problem states that x is the number of units demanded and that D(x) is the demand function. I think it's mistakenly written by my professor.. I got your point and thanks a lot.. :)
$endgroup$
– marg_ocruz
May 24 '15 at 9:21
$begingroup$
the problem states that x is the number of units demanded and that D(x) is the demand function. I think it's mistakenly written by my professor.. I got your point and thanks a lot.. :)
$endgroup$
– marg_ocruz
May 24 '15 at 9:21
add a comment |
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