What is the demand function p(x)?
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0
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The marginal revenue of a certain commodity is $R^1(x)=-3x^2+4x+32$
where $x$ is the level of production in thousands. Assume $R(0)=0$ Find $R(x)$. What is the demand function of $p(x)$?
I took the integral and got $R(x)=-x^3+2x^2+32x$
I'm not sure how to find the demand function though.
Any help would be appreciated. Thanks
calculus optimization
asked Dec 12 '14 at 21:53
user200779
12
add a comment |
up vote
0
down vote
favorite
The marginal revenue of a certain commodity is $R^1(x)=-3x^2+4x+32$
where $x$ is the level of production in thousands. Assume $R(0)=0$ Find $R(x)$. What is the demand function of $p(x)$?
I took the integral and got $R(x)=-x^3+2x^2+32x$
I'm not sure how to find the demand function though.
Any help would be appreciated. Thanks
calculus optimization
asked Dec 12 '14 at 21:53
user200779
12
The first term in your integral should be $x^3$
– Ross Millikan
Dec 12 '14 at 22:01
Yes, that was my mistake when copying the answer to here.
– user200779
Dec 12 '14 at 22:06
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The marginal revenue of a certain commodity is $R^1(x)=-3x^2+4x+32$
where $x$ is the level of production in thousands. Assume $R(0)=0$ Find $R(x)$. What is the demand function of $p(x)$?
I took the integral and got $R(x)=-x^3+2x^2+32x$
I'm not sure how to find the demand function though.
Any help would be appreciated. Thanks
calculus optimization
asked Dec 12 '14 at 21:53
user200779
12
The marginal revenue of a certain commodity is $R^1(x)=-3x^2+4x+32$
where $x$ is the level of production in thousands. Assume $R(0)=0$ Find $R(x)$. What is the demand function of $p(x)$?
I took the integral and got $R(x)=-x^3+2x^2+32x$
I'm not sure how to find the demand function though.
Any help would be appreciated. Thanks
calculus optimization
calculus optimization
asked Dec 12 '14 at 21:53
user200779
12
asked Dec 12 '14 at 21:53
user200779
12
edited Dec 12 '14 at 22:07
asked Dec 12 '14 at 21:53
user200779
12
asked Dec 12 '14 at 21:53
user200779
12
asked Dec 12 '14 at 21:53
user200779
12
12
The first term in your integral should be $x^3$
– Ross Millikan
Dec 12 '14 at 22:01
Yes, that was my mistake when copying the answer to here.
– user200779
Dec 12 '14 at 22:06
add a comment |
The first term in your integral should be $x^3$
– Ross Millikan
Dec 12 '14 at 22:01
Yes, that was my mistake when copying the answer to here.
– user200779
Dec 12 '14 at 22:06
The first term in your integral should be $x^3$
– Ross Millikan
Dec 12 '14 at 22:01
The first term in your integral should be $x^3$
– Ross Millikan
Dec 12 '14 at 22:01
Yes, that was my mistake when copying the answer to here.
– user200779
Dec 12 '14 at 22:06
Yes, that was my mistake when copying the answer to here.
– user200779
Dec 12 '14 at 22:06
add a comment |
1 Answer
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0
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Hint: The revenue function is $R(x)=p(x)cdot x$ Thus you can calculate $p(x)$.
answered Dec 12 '14 at 22:16
callculus
17.7k31427
So, do I just factor out the x to get p(x). It ends up being p(x)=-x^2+2x+32 or am I off?
– user200779
Dec 12 '14 at 22:30
Yes. Note that the condition $R(0)=0$ is satisfied, because $R(0)=-0^3+2cdot 0^2+32cdot 0=0$.
– callculus
Dec 12 '14 at 22:34
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
0
down vote
Hint: The revenue function is $R(x)=p(x)cdot x$ Thus you can calculate $p(x)$.
answered Dec 12 '14 at 22:16
callculus
17.7k31427
So, do I just factor out the x to get p(x). It ends up being p(x)=-x^2+2x+32 or am I off?
– user200779
Dec 12 '14 at 22:30
Yes. Note that the condition $R(0)=0$ is satisfied, because $R(0)=-0^3+2cdot 0^2+32cdot 0=0$.
– callculus
Dec 12 '14 at 22:34
add a comment |
up vote
0
down vote
Hint: The revenue function is $R(x)=p(x)cdot x$ Thus you can calculate $p(x)$.
answered Dec 12 '14 at 22:16
callculus
17.7k31427
So, do I just factor out the x to get p(x). It ends up being p(x)=-x^2+2x+32 or am I off?
– user200779
Dec 12 '14 at 22:30
Yes. Note that the condition $R(0)=0$ is satisfied, because $R(0)=-0^3+2cdot 0^2+32cdot 0=0$.
– callculus
Dec 12 '14 at 22:34
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint: The revenue function is $R(x)=p(x)cdot x$ Thus you can calculate $p(x)$.
answered Dec 12 '14 at 22:16
callculus
17.7k31427
Hint: The revenue function is $R(x)=p(x)cdot x$ Thus you can calculate $p(x)$.
answered Dec 12 '14 at 22:16
callculus
17.7k31427
answered Dec 12 '14 at 22:16
callculus
17.7k31427
answered Dec 12 '14 at 22:16
callculus
17.7k31427
answered Dec 12 '14 at 22:16
callculus
17.7k31427
17.7k31427
So, do I just factor out the x to get p(x). It ends up being p(x)=-x^2+2x+32 or am I off?
– user200779
Dec 12 '14 at 22:30
Yes. Note that the condition $R(0)=0$ is satisfied, because $R(0)=-0^3+2cdot 0^2+32cdot 0=0$.
– callculus
Dec 12 '14 at 22:34
add a comment |
So, do I just factor out the x to get p(x). It ends up being p(x)=-x^2+2x+32 or am I off?
– user200779
Dec 12 '14 at 22:30
Yes. Note that the condition $R(0)=0$ is satisfied, because $R(0)=-0^3+2cdot 0^2+32cdot 0=0$.
– callculus
Dec 12 '14 at 22:34
So, do I just factor out the x to get p(x). It ends up being p(x)=-x^2+2x+32 or am I off?
– user200779
Dec 12 '14 at 22:30
So, do I just factor out the x to get p(x). It ends up being p(x)=-x^2+2x+32 or am I off?
– user200779
Dec 12 '14 at 22:30
Yes. Note that the condition $R(0)=0$ is satisfied, because $R(0)=-0^3+2cdot 0^2+32cdot 0=0$.
– callculus
Dec 12 '14 at 22:34
Yes. Note that the condition $R(0)=0$ is satisfied, because $R(0)=-0^3+2cdot 0^2+32cdot 0=0$.
– callculus
Dec 12 '14 at 22:34
add a comment |
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The first term in your integral should be $x^3$
– Ross Millikan
Dec 12 '14 at 22:01
Yes, that was my mistake when copying the answer to here.
– user200779
Dec 12 '14 at 22:06