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Demand for your tie-dyed T-shirts is given by the formula q = 510 − 90p0.5 where q is the number of T-shirts you can sell each month at a price of p dollars. If you currently sell T-shirts for $15 each and you raise your price by $2 per month, how fast will the demand drop? (Round your answer to the nearest whole number.)

2 Answers

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Final answer:

The demand for tie-dyed T-shirts will drop by approximately 23 T-shirts when the price is increased from $15 to $17. This calculation is based on the function q = 510 − 90
p^(0.5), which represents the demand in relation to the price.

Step-by-step explanation:

The student is asking how fast the demand for tie-dyed T-shirts will drop if the price increases by $2 when the demand is represented by the function q = 510 − 90
p^(0.5). Since the current price is $15, we need to calculate the demand at both $15 and $17 to find the rate of change in demand when the price is raised by $2.

To find the demand at the current price of $15, substitute p = 15 into the demand function:
q = 510 − 90
(15)^(0.5)
q = 510 − 90(3.87)
q = 510 − 348.3
q = 161.7

To find the demand when the price is $17, substitute p = 17 into the demand function:
q = 510 − 90
(17)^(0.5)
q = 510 − 90(4.12)
q = 510 − 371
q = 139

The change in demand is the difference between the two quantities:
Δq = 161.7 - 139 = 22.7
The demand drops by approximately 23 T-shirts when the price increases by $2.

User Roy Goode
by
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4 votes

Answer:

The demand will drop approximately 23 T-shirt per month.

Step-by-step explanation:

Consider the provided formula.


q = 510-90p^(0.5)

You currently sell T-shirts for $15 each and you raise your price by $2 per month,

That means p=$15 and
(dp)/(dt)=2

We need to find how fast demand will drop.

So differentiate the above function with respect to time.


(dq)/(dt) =0-45p^(0.5-1)(dp)/(dt)


(dq)/(dt) =-45p^(-0.5)(dp)/(dt)

Substitute the respective values as shown:


(dq)/(dt) =-45(15)^(-0.5)(2)


(dq)/(dt) \approx-23.24

Hence, the demand will drop approximately 23 T-shirt per month.

User Thomas Gassmann
by
5.8k points