Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
The demand function for a product is given by p = 10,000(1 − (5/5 + e^−0.001x)) where p is the price per unit (in dollars) and...Asked by steve
The demand function for a product is given by
p = 10,000(1 − (5/5 + e^−0.001x))
where p is the price per unit (in dollars) and x is the number of units sold. Find the numbers of units sold for prices of p = $1000 and p = $1500.(Round your answers to the nearest integer.)
(a) p = $1000____
(b) p = $1500____
p = 10,000(1 − (5/5 + e^−0.001x))
where p is the price per unit (in dollars) and x is the number of units sold. Find the numbers of units sold for prices of p = $1000 and p = $1500.(Round your answers to the nearest integer.)
(a) p = $1000____
(b) p = $1500____
Answers
Answered by
Steve
what's the problem? Set p=1000 and solve for x:
1000 = 10,000(1 − (5/5 + e^−0.001x))
0.1 = 1 − (5/5 + e^−0.001x)
0.9 = 5/5 + e^−0.001x
I suspect a typo,for 5/5 makes little sense. However, moving right along,
-0.1 = e^-.001x
Now we know there's a boo-boo, since the exponential function is never negative.
I suggest you fix the error, then continue on with the solution. When you get to the point we reached above, take log of both sides.
1000 = 10,000(1 − (5/5 + e^−0.001x))
0.1 = 1 − (5/5 + e^−0.001x)
0.9 = 5/5 + e^−0.001x
I suspect a typo,for 5/5 makes little sense. However, moving right along,
-0.1 = e^-.001x
Now we know there's a boo-boo, since the exponential function is never negative.
I suggest you fix the error, then continue on with the solution. When you get to the point we reached above, take log of both sides.
Answered by
MathMate
p = 10,000(1 − (5/5 + e^−0.001x))
for p=1000,
1000=10000(1 − (5/5 + e^−0.001x))
5/(5+e^(-0.001x)=0.9
e^(-0.001x)=5/0.9-5
take log
-0.001x = log(5/0.9-5)
x=-1000 log(5/0.9-5)
=587.8,
say 588
(b) can be done in a similar way.
for p=1000,
1000=10000(1 − (5/5 + e^−0.001x))
5/(5+e^(-0.001x)=0.9
e^(-0.001x)=5/0.9-5
take log
-0.001x = log(5/0.9-5)
x=-1000 log(5/0.9-5)
=587.8,
say 588
(b) can be done in a similar way.
Answered by
MathMate
Steve is right.
Whenever you transpose a fraction from a book to a single line to post, you need to insert the implicit parentheses. In this case, I believe it is
1000 = 10,000(1 − (5/[5 + e^−0.001x]))
Whenever you transpose a fraction from a book to a single line to post, you need to insert the implicit parentheses. In this case, I believe it is
1000 = 10,000(1 − (5/[5 + e^−0.001x]))
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.