# Word problem: Cost function, Revenue Function, Profit function.

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• September 18th 2011, 01:47 PM
hollylovespunk
Word problem: Cost function, Revenue Function, Profit function.
Hey guys I have a word problem and I was wondering if you could check my work. A factory can product 1,000 bottles per hour. The machine (with setup) cost \$3,000,000, whilst each bottle costs .5 cents (labor,materials, etc.) to make. The machine runs 10 hours per day. If each bottle sells for .65 cents, express the cost, revenue and profit for running the factory for X days. After how many days will the factory break even?

Cost function, C(x) = 3,000,000 + .05(10000X)
= 3,000,000 + 5000X
Revenue function, R(x) = .65(10000X)
= 6500X
Profit function, P(x) = 6500X - 3000000 + 5000X
= 11500X - 3000000
Break Even: 3000000 = 11500X
So in approximately 261 days the factory will break even.

Am I right?
• September 18th 2011, 01:50 PM
skeeter
Re: Word problem: Cost function, Revenue Function, Profit function.
Quote:

Originally Posted by hollylovespunk
Hey guys I have a word problem and I was wondering if you could check my work. A factory can product 1,000 bottles per hour. The machine (with setup) cost \$3,000,000, whilst each bottle costs .5 cents (labor,materials, etc.) to make. The machine runs 10 hours per day. If each bottle sells for .65 cents, express the cost, revenue and profit for running the factory for X days. After how many days will the factory break even?

Cost function, C(x) = 3,000,000 + .05(10000X)
= 3,000,000 + 5000X
Revenue function, R(x) = .65(10000X)
= 6500X
Profit function, P(x) = 6500X - 3000000 + 5000X
= 11500X - 3000000
Break Even: 3000000 = 11500X
So in approximately 261 days the factory will break even.

Am I right?

no ...

P(x) = R(x) - C(x) = 6500x - (3,000,000 + 5000x)

try again
• September 18th 2011, 02:06 PM
hollylovespunk
Re: Word problem: Cost function, Revenue Function, Profit function.
So its 1500X. So the factory would break even after 2000 days?
• September 18th 2011, 02:18 PM
skeeter
Re: Word problem: Cost function, Revenue Function, Profit function.
after looking again, I notice you've also made one significant arithmetic error in your original function.

$.05(10000x) \ne 5000x$
• September 18th 2011, 02:22 PM
hollylovespunk
Re: Word problem: Cost function, Revenue Function, Profit function.
Why does 0.5 multiplied by 10000X not equal 5000X?
• September 18th 2011, 02:24 PM
skeeter
Re: Word problem: Cost function, Revenue Function, Profit function.
Quote:

Originally Posted by hollylovespunk
Why does 0.5 multiplied by 10000X not equal 5000X?

0.5 cents = 0.005 dollars
• September 18th 2011, 02:26 PM
hollylovespunk
Re: Word problem: Cost function, Revenue Function, Profit function.
(Headbang) Thanks for pointing that out. How does the set up look?
• September 18th 2011, 02:35 PM
skeeter
Re: Word problem: Cost function, Revenue Function, Profit function.
Quote:

Originally Posted by hollylovespunk
(Headbang)Thanks. How does the rest look?f

make the correction, solve the problem and then I'll respond.
• September 18th 2011, 02:39 PM
hollylovespunk
Re: Word problem: Cost function, Revenue Function, Profit function.
C(x) = 3000000 + 500X

R(x) = 6500X

P(x) = 6500X - (3000000 + 500X)
= 6000X - 3000000
Break Even: 6000X = 3000000

The factory would break even after 500 days
• September 18th 2011, 03:00 PM
skeeter
Re: Word problem: Cost function, Revenue Function, Profit function.
Quote:

The machine (with setup) cost \$3,000,000, whilst each bottle costs .5 cents (labor,materials, etc.) to make. The machine runs 10 hours per day. If each bottle sells for .65 cents, express the cost, revenue and profit for running the factory for X days.
0.5 cents = 0.005 dollars ... 0.65 cents = 0.0065 dollars

that means a profit of only 0.0015 dollars per bottle x 10000 = \$15.00 per day

going to take an unreasonably long time to break even with 3 million dollars

something's not right with the problem statement ... recheck the bottle cost and selling prices.
• September 18th 2011, 03:04 PM
hollylovespunk
Re: Word problem: Cost function, Revenue Function, Profit function.
I am sorry. I just realized what I did. The problem says 5 cents. I entered it incorrectly as 0.5 cents (Headbang)
• September 18th 2011, 03:05 PM
skeeter
Re: Word problem: Cost function, Revenue Function, Profit function.
Quote:

Originally Posted by hollylovespunk
I am sorry. I just realized what I did. The problem says 5 cents. I entered it incorrectly as 0.5 cents (Headbang)

how about the selling price ... 0.65 cents, 6.5 cents, or 65 cents?
• September 19th 2011, 04:07 PM
hollylovespunk
Re: Word problem: Cost function, Revenue Function, Profit function.
The problem says 65 cents (Headbang) I should really work smart and not hard (Wink)
• September 25th 2011, 03:05 PM
hollylovespunk
Re: Word problem: Cost function, Revenue Function, Profit function.
Well here is my solution:

C(x) = 3000000 + (10000x) .05
= 3000000 + 500x

R(x) = .65(10000x)
= 6500x

P(x) = R(x) - C(x)
= 6500x - 3000000 - 500x
= 6000x - 3000000

Break even: R(x) = C(x)
6500x = 3000000 + 500x
6000x = 3000000
Break even after 500 days.

Looks good?
• September 25th 2011, 04:21 PM
skeeter
Re: Word problem: Cost function, Revenue Function, Profit function.
Quote:

Originally Posted by hollylovespunk
Well here is my solution:

C(x) = 3000000 + (10000x) .05
= 3000000 + 500x

R(x) = .65(10000x)
= 6500x

P(x) = R(x) - C(x)
= 6500x - 3000000 - 500x
= 6000x - 3000000

Break even: R(x) = C(x)
6500x = 3000000 + 500x
6000x = 3000000
Break even after 500 days.

Looks good?

that's fine.
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