Number of Bicycles

[harpazo]

Suppose that the government imposes a 1000 dollars per day tax on the bicycle manufacturer so that the daily cost C of manufacturing x bicycles is now given by C(x) = 80x + 6000.
Now the average daily cost C (bar) is given by C(bar) of x = (80x + 6000)/x. How many bicycles must be produced each day for the average cost to be no more than 100 dollars?

Must I evaluate C(bar) of (100) to find the number of bicycles? See attachment for C(bar) of x.

 

[MarkFL]

You recently posted this same question here:

Average Cost 2

Suppose that the government imposes a $1000 per day tax on the bicycle manufacturer so that the daily cost C of manufacturing x bicycles is now given by C(x) = 80x + 6000. Now the average daily cost is given by (x). How many bicycles must be produced each day for the average cost to be no more...

www.freemathhelp.com

 

[Deleted member 4993]

Suppose that the government imposes a 1000 dollars per day tax on the bicycle manufacturer so that the daily cost C of manufacturing x bicycles is now given by C(x) = 80x + 6000.
Now the average daily cost C (bar) is given by C(bar) of x = (80x + 6000)/x. How many bicycles must be produced each day for the average cost to be no more than 100 dollars?

Must I evaluate C(bar) of (100) to find the number of bicycles? See attachment for C(bar) of x.

View attachment 12439

What do you need to find? → Number of bicycles = x

Here x is not equal to 100

Solve for x such that:

C(bar) = (80*x + 6000)/x ≤ 100

 

[harpazo]

You recently posted this same question here:

Average Cost 2

Suppose that the government imposes a $1000 per day tax on the bicycle manufacturer so that the daily cost C of manufacturing x bicycles is now given by C(x) = 80x + 6000. Now the average daily cost is given by (x). How many bicycles must be produced each day for the average cost to be no more...

www.freemathhelp.com

Isn't the question slightly different? There is no C(bar) in the link you provided.

 

[harpazo]

What do you need to find? → Number of bicycles = x

Here x is not equal to 100

Solve for x such that:

C(bar) = (80*x + 6000)/x ≤ 100

What do you need to find? → Number of bicycles = x

Here x is not equal to 100

Solve for x such that:

C(bar) = (80*x + 6000)/x ≤ 100

C(bar) = (80x + 6000)/x ≤ 100

Multiply both sides by x we get

80x + 6000 ≤ 100x

80x - 100x ≤ -6000

-20x ≤ -6000

x ≥ -6000/-20

x ≥ 300

So, 300 bicycles must be produced each day for the average cost to be no more than 100 dollars.

 

[MarkFL]

Isn't the question slightly different? There is no C(bar) in the link you provided.

The wording may be slightly different, but it is the same exact problem, is it not?

 

[Dr.Peterson]

Isn't the question slightly different? There is no C(bar) in the link you provided.

In the last post in that thread, I showed evidence that there was a [MATH]\overline{C}[/MATH] in the original of your problem, and that it didn't change the meaning if you overlooked the nonsense of saying (x).

In any case, your work both there and here is correct (though what you did this time is better), so it really doesn't matter. And since no one in either thread has yet told you that, it's not entirely silly for you to have asked again.

By the way, you can check by first putting in 300 for x and seeing that the average cost is $100, and then trying a larger number and confirming that it is less than $100.

 

[MarkFL]

I guess my issue is that the same exact question was posted twice, and there was yet another question posted prior to both that was virtually identical, and for which a complete solution was provided. I hate to see our helpers laboring through so much redundancy, and is the primary reason I have repeatedly asked that one question of a particular type be initially posted, and then an attempt be made to apply the help given for that initial problem to all virtually identical subsequent problems.

 

[Dr.Peterson]

I agree. The thread referred to here went nowhere, so it was never properly answered, but the other should have sufficed to cover the issue. (I did say "not entirely silly".)

The question, for which I don't envy moderators, is how to help people learn to use a site well.