the price point that will maximize the revenue, and; what the maximum revenue is
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The other day, as you are burying yourself in the sea of data, you find out that at $500$, your corporation sells on average $50,000$ Strawberry Phones, the corporation’s flagship product, monthly. Additionally, for every $50$ increase in the price of a phone, the sales decrease by $1,000$ phones. Based on this information, you are very interested in finding
the price point that will maximize the revenue, and;
what the maximum revenue is
calculus
edited Nov 26 at 21:01
Larry
1,2822722
asked Nov 26 at 20:51
Jennifer Gonzaga
12
add a comment |
up vote
-1
down vote
favorite
The other day, as you are burying yourself in the sea of data, you find out that at $500$, your corporation sells on average $50,000$ Strawberry Phones, the corporation’s flagship product, monthly. Additionally, for every $50$ increase in the price of a phone, the sales decrease by $1,000$ phones. Based on this information, you are very interested in finding
the price point that will maximize the revenue, and;
what the maximum revenue is
calculus
edited Nov 26 at 21:01
Larry
1,2822722
asked Nov 26 at 20:51
Jennifer Gonzaga
12
We use dollar signs to set off MathJax. You either need to escape them with backslashes or remove them. Then, what have you tried? Where are you stuck?
– Ross Millikan
Nov 26 at 20:55
I tried q(p)=49,480p+b ... Q(50)=49,480(50)+b=1,000
– Jennifer Gonzaga
Nov 26 at 21:04
1
Please define your variables. Presumably $p$ is price and $q$ or $Q$ (they are not the same) is quantity. What is $b$? You should find the revenue as a function of one variable, then take the derivative and set to zero. The equation that links quantity to price is how you get rid of the second variable.
– Ross Millikan
Nov 26 at 21:10
We are told that if $500 + 50x$ phones are sold, then the number of sales is $50000 - 1000x$. Can you write a revenue equation using that information?
– N. F. Taussig
Nov 27 at 13:35
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
The other day, as you are burying yourself in the sea of data, you find out that at $500$, your corporation sells on average $50,000$ Strawberry Phones, the corporation’s flagship product, monthly. Additionally, for every $50$ increase in the price of a phone, the sales decrease by $1,000$ phones. Based on this information, you are very interested in finding
the price point that will maximize the revenue, and;
what the maximum revenue is
calculus
edited Nov 26 at 21:01
Larry
1,2822722
asked Nov 26 at 20:51
Jennifer Gonzaga
12
The other day, as you are burying yourself in the sea of data, you find out that at $500$, your corporation sells on average $50,000$ Strawberry Phones, the corporation’s flagship product, monthly. Additionally, for every $50$ increase in the price of a phone, the sales decrease by $1,000$ phones. Based on this information, you are very interested in finding
the price point that will maximize the revenue, and;
what the maximum revenue is
calculus
calculus
edited Nov 26 at 21:01
Larry
1,2822722
asked Nov 26 at 20:51
Jennifer Gonzaga
12
edited Nov 26 at 21:01
Larry
1,2822722
asked Nov 26 at 20:51
Jennifer Gonzaga
12
edited Nov 26 at 21:01
Larry
1,2822722
edited Nov 26 at 21:01
Larry
1,2822722
edited Nov 26 at 21:01
Larry
1,2822722
1,2822722
asked Nov 26 at 20:51
Jennifer Gonzaga
12
asked Nov 26 at 20:51
Jennifer Gonzaga
12
asked Nov 26 at 20:51
Jennifer Gonzaga
12
12
We use dollar signs to set off MathJax. You either need to escape them with backslashes or remove them. Then, what have you tried? Where are you stuck?
– Ross Millikan
Nov 26 at 20:55
I tried q(p)=49,480p+b ... Q(50)=49,480(50)+b=1,000
– Jennifer Gonzaga
Nov 26 at 21:04
1
Please define your variables. Presumably $p$ is price and $q$ or $Q$ (they are not the same) is quantity. What is $b$? You should find the revenue as a function of one variable, then take the derivative and set to zero. The equation that links quantity to price is how you get rid of the second variable.
– Ross Millikan
Nov 26 at 21:10
We are told that if $500 + 50x$ phones are sold, then the number of sales is $50000 - 1000x$. Can you write a revenue equation using that information?
– N. F. Taussig
Nov 27 at 13:35
add a comment |
We use dollar signs to set off MathJax. You either need to escape them with backslashes or remove them. Then, what have you tried? Where are you stuck?
– Ross Millikan
Nov 26 at 20:55
I tried q(p)=49,480p+b ... Q(50)=49,480(50)+b=1,000
– Jennifer Gonzaga
Nov 26 at 21:04
1
Please define your variables. Presumably $p$ is price and $q$ or $Q$ (they are not the same) is quantity. What is $b$? You should find the revenue as a function of one variable, then take the derivative and set to zero. The equation that links quantity to price is how you get rid of the second variable.
– Ross Millikan
Nov 26 at 21:10
We are told that if $500 + 50x$ phones are sold, then the number of sales is $50000 - 1000x$. Can you write a revenue equation using that information?
– N. F. Taussig
Nov 27 at 13:35
We use dollar signs to set off MathJax. You either need to escape them with backslashes or remove them. Then, what have you tried? Where are you stuck?
– Ross Millikan
Nov 26 at 20:55
We use dollar signs to set off MathJax. You either need to escape them with backslashes or remove them. Then, what have you tried? Where are you stuck?
– Ross Millikan
Nov 26 at 20:55
I tried q(p)=49,480p+b ... Q(50)=49,480(50)+b=1,000
– Jennifer Gonzaga
Nov 26 at 21:04
I tried q(p)=49,480p+b ... Q(50)=49,480(50)+b=1,000
– Jennifer Gonzaga
Nov 26 at 21:04
1
1
Please define your variables. Presumably $p$ is price and $q$ or $Q$ (they are not the same) is quantity. What is $b$? You should find the revenue as a function of one variable, then take the derivative and set to zero. The equation that links quantity to price is how you get rid of the second variable.
– Ross Millikan
Nov 26 at 21:10
Please define your variables. Presumably $p$ is price and $q$ or $Q$ (they are not the same) is quantity. What is $b$? You should find the revenue as a function of one variable, then take the derivative and set to zero. The equation that links quantity to price is how you get rid of the second variable.
– Ross Millikan
Nov 26 at 21:10
We are told that if $500 + 50x$ phones are sold, then the number of sales is $50000 - 1000x$. Can you write a revenue equation using that information?
– N. F. Taussig
Nov 27 at 13:35
We are told that if $500 + 50x$ phones are sold, then the number of sales is $50000 - 1000x$. Can you write a revenue equation using that information?
– N. F. Taussig
Nov 27 at 13:35
add a comment |
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We use dollar signs to set off MathJax. You either need to escape them with backslashes or remove them. Then, what have you tried? Where are you stuck?
– Ross Millikan
Nov 26 at 20:55
I tried q(p)=49,480p+b ... Q(50)=49,480(50)+b=1,000
– Jennifer Gonzaga
Nov 26 at 21:04
1
Please define your variables. Presumably $p$ is price and $q$ or $Q$ (they are not the same) is quantity. What is $b$? You should find the revenue as a function of one variable, then take the derivative and set to zero. The equation that links quantity to price is how you get rid of the second variable.
– Ross Millikan
Nov 26 at 21:10
We are told that if $500 + 50x$ phones are sold, then the number of sales is $50000 - 1000x$. Can you write a revenue equation using that information?
– N. F. Taussig
Nov 27 at 13:35