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 _____units
(b) p = $1500 _____units
To find the number of units sold for a given price, we need to solve the demand function for x when p is equal to the given price.
(a) For p = $1000:
Substitute p = $1000 into the demand function and solve for x:
$1000 = 10,000 [1 − (5/5 + e^−0.001x)]
Dividing both sides of the equation by 10,000:
0.1 = 1 − (5/5 + e^−0.001x)
Using the property a/(b+c) = a/b + a/c:
0.1 = 1 − (1 + e^−0.001x)
Combining like terms:
0.1 = 1 - 1 - e^−0.001x
0.1 = - e^−0.001x
Taking the natural logarithm of both sides:
ln(0.1) = ln(− e^−0.001x)
Using the property ln(e^a) = a:
ln(0.1) = −0.001x
Solving for x:
x = ln(0.1) / −0.001
Using a calculator, we find:
x ≈ 2296
Hence, for p = $1000, approximately 2296 units are sold.
(b) For p = $1500:
Follow the same steps as above, but substitute p = $1500 into the demand function:
$1500 = 10,000 [1 − (5/5 + e^−0.001x)]
Dividing both sides by 10,000:
0.15 = 1 − (5/5 + e^−0.001x)
Using the property a/(b+c) = a/b + a/c:
0.15 = 1 − (1 + e^−0.001x)
Combining like terms:
0.15 = 1 - 1 - e^−0.001x
0.15 = - e^−0.001x
Taking the natural logarithm of both sides:
ln(0.15) = ln(− e^−0.001x)
Using the property ln(e^a) = a:
ln(0.15) = −0.001x
Solving for x:
x = ln(0.15) / −0.001
Using a calculator, we find:
x ≈ 3470
Hence, for p = $1500, approximately 3470 units are sold.