Quadratic Equation for Profits

Caity · #1 $Old$ January 5th, 2009, 08:42 AM

A retailer spent $48 to purchase a number of special mugs. Two of them were broken in the store, but by selling each of the remaining mugs for $3 above the original cost per mug, she made a total profit of $22.

Construct an equation that will allow us to solve for the number of mugs, denoted by n, that were originally purchased.

a) If the price for n mugs is $48, how can we express the cost per mug?

for this I got $\frac{48}{n}$

b) We are told that the retailer, in order to make a specific profit, sells the remaining mugs for $3 more than the original cost per mug. Using this information, and your answer to step a), how can we represent the number of mugs that are available for sale?

for this I got $\frac{48}{n} +3$

c) Since two of the mugs were broken in the store, how can we represent the number of mugs that are available for sale?

for this I got $n - 2$ but not sure about it....

d) Use your answers from steps b) and c) to construct and expression for the amount of money made on the sale of the remaining mugs.

for this I have $n(\frac{48}{n} +3) - 6$

e) The profit on the mugs can be found by subtracting the cost fo the mugs from the amount of money made on the sale of the mugs. Write an equation that states this relationship between the money made, the cost, and the profit.

dont know how to do this one...

The real problem is

Rewrite the equation that you wrote up in step e) in the standard form of a quadratic equation and show your work:

Problem

Solve the quadratic equation an answer the following questions.
1. How many mugs were purchased?
2. What was the origina cost per mug?
3. What profit was made on each mug?
4. When solving a quadratic equation, there are normally two solutions. Why were you able to discard one of the solutions in this case?

Last_Singularity · #2 $Old$ January 5th, 2009, 10:34 AM

Quote:

Originally Posted by Caity $View Post$

A retailer spent $48 to purchase a number of special mugs. Two of them were broken in the store, but by selling each of the remaining mugs for $3 above the original cost per mug, she made a total profit of $22.

Construct an equation that will allow us to solve for the number of mugs, denoted by n, that were originally purchased.

a) If the price for n mugs is $48, how can we express the cost per mug?

for this I got $\frac{48}{n}$

b) We are told that the retailer, in order to make a specific profit, sells the remaining mugs for $3 more than the original cost per mug. Using this information, and your answer to step a), how can we represent the number of mugs that are available for sale?

for this I got $\frac{48}{n} +3$

c) Since two of the mugs were broken in the store, how can we represent the number of mugs that are available for sale?

for this I got $n - 2$ but not sure about it....

d) Use your answers from steps b) and c) to construct and expression for the amount of money made on the sale of the remaining mugs.

for this I have $n(\frac{48}{n} +3) - 6$

e) The profit on the mugs can be found by subtracting the cost fo the mugs from the amount of money made on the sale of the mugs. Write an equation that states this relationship between the money made, the cost, and the profit.

dont know how to do this one...

The real problem is

Rewrite the equation that you wrote up in step e) in the standard form of a quadratic equation and show your work:

Problem

Solve the quadratic equation an answer the following questions.
1. How many mugs were purchased?
2. What was the origina cost per mug?
3. What profit was made on each mug?
4. When solving a quadratic equation, there are normally two solutions. Why were you able to discard one of the solutions in this case?

Originally, the retailer spent $48, on an unknown number of mugs. Call this number $n$ . Therefore, the cost of buying $n$ mugs, for a total of $48, gives a unit price for $48/n$ (units of dollars per mug)

Two mugs broke, so she has $n-2$ mugs left over for sell.

The unit price is $3 higher than the original one, so the new unit price is $48/n+3$

So the revenue of selling these $n-2$ mugs at $48/n+3$ each is clearly $(n-2)(48/n+3)$

Profit is given as the difference between revenue and costs, or profit = revenue - cost.

So if the profit is $22, your equation becomes:
$22 = (n-2)(48/n+3) - 48$

This is a quadratic equation - you can now solve for $n$ . You should be able to solve the remainder of the problem.

And the last part will become obvious once you solve the equation for the values of $n$ : one of the roots will be negative. So of course you toss this one out because buying a negative number of mugs does not really make sense in the real world.

HallsofIvy · #3 $Old$ January 6th, 2009, 05:59 AM

Quote:

Originally Posted by Caity $View Post$

A retailer spent $48 to purchase a number of special mugs. Two of them were broken in the store, but by selling each of the remaining mugs for $3 above the original cost per mug, she made a total profit of $22.

Construct an equation that will allow us to solve for the number of mugs, denoted by n, that were originally purchased.

a) If the price for n mugs is $48, how can we express the cost per mug?

for this I got $\frac{48}{n}$

Yes, that is correct.

Quote:

b) We are told that the retailer, in order to make a specific profit, sells the remaining mugs for $3 more than the original cost per mug. Using this information, and your answer to step a), how can we represent the number of mugs that are available for sale?

for this I got $\frac{48}{n} +3$

You said in a that $\frac{48}{n}$ was a cost per mug and the only "3" in the problem is the selling price $3 per mug. So " $\frac{48}{n}+ 3$ must have units of "dollars per mug". That cannot be the number of mugs!

But it doesn't make sense that the question asked here is exactly the same as the question asked in (c)! Did you mis-copy the question? If the question was "what price did he sell the mugs for", then your answer is correct.

Quote:

c) Since two of the mugs were broken in the store, how can we represent the number of mugs that are available for sale?

for this I got $n - 2$ but not sure about it....

Well, of course! If you start with n things and "remove" 2 of them, you have n-2 left.

Quote:

d) Use your answers from steps b) and c) to construct and expression for the amount of money made on the sale of the remaining mugs.

for this I have $n(\frac{48}{n} +3) - 6$

You answer to (b) was $\frac{48}{n}+ 3$ dollars per mug and to (c) was n- 2 mugs. If you multiply those you get "dollars" as you want. But $(n-2)(\frac{48}{n}+ 3)$ is NOT $n(\frac{48}{n}+ 3)- 6$ . You have to multiply the "-2" by $\frac{48}{n}$ as well as by 3. And you have to multiply the "n" by 3 as well as [LaTeX Error: Syntax error].
$(n-2)(\frac{48}{n}+ 3)= n(\frac{48}{n})+ 3n- 2\frac{48}{n}- 6$
$= 42+ 3n- \frac{96}{n}$

[/quote]e) The profit on the mugs can be found by subtracting the cost fo the mugs from the amount of money made on the sale of the mugs. Write an equation that states this relationship between the money made, the cost, and the profit.

dont know how to do this one...
The mugs cost a total of $48 so the profit, $12, is the money in (d) minus 48:
$42+ 3n- 2\frac{48}{n}- 48= 12$
$3n- \frac{96}{n}- 18= 0$

The real problem is

Rewrite the equation that you wrote up in step e) in the standard form of a quadratic equation and show your work:[/quote]
Since the equation above involves an n in the denominator, multiply every term in the equation by n:
$3n^2- 18n- 96= 0$
or, dividing through by 3,
$n^2- 6n- 32= 0$

[/quote]Problem

Solve the quadratic equation an answer the following questions.
1. How many mugs were purchased?
2. What was the origina cost per mug?
3. What profit was made on each mug?
4. When solving a quadratic equation, there are normally two solutions. Why were you able to discard one of the solutions in this case?[/quote]

Now that you have the right equation, try these yourself.