Math Problem Statement

The monthly sales of Sunny Electronics' new sound system are given by q(t) = 2,000t − 100t^2 units per month, t months after its introduction. The price Sunny charges is p(t) = 1,000 − t^2 dollars per sound system, t months after introduction. Find the rate of change of monthly sales. q'(t) =

Find the rate of change of the price. p'(t) =

Find the rate of change of monthly revenue 5 months after the introduction of the sound system. $ per month

Interpret your answer. When t = , the revenue is increasing at a rate of $ per month.

Solution

Let's break this down step by step.

1. Find the rate of change of monthly sales: q(t)q'(t)

The monthly sales function is given by: q(t)=2000t100t2q(t) = 2000t - 100t^2

To find the rate of change of sales, we differentiate q(t)q(t) with respect to tt:

q(t)=ddt(2000t100t2)q'(t) = \frac{d}{dt}(2000t - 100t^2) q(t)=2000200tq'(t) = 2000 - 200t

So, the rate of change of monthly sales is: q(t)=2000200tq'(t) = 2000 - 200t

2. Find the rate of change of the price: p(t)p'(t)

The price function is given by: p(t)=1000t2p(t) = 1000 - t^2

We differentiate p(t)p(t) with respect to tt:

p(t)=ddt(1000t2)p'(t) = \frac{d}{dt}(1000 - t^2) p(t)=2tp'(t) = -2t

So, the rate of change of the price is: p(t)=2tp'(t) = -2t

3. Find the rate of change of monthly revenue 5 months after the introduction of the sound system

Revenue R(t)R(t) is the product of price p(t)p(t) and quantity sold q(t)q(t):

R(t)=p(t)×q(t)R(t) = p(t) \times q(t)

Substitute the expressions for p(t)p(t) and q(t)q(t):

R(t)=(1000t2)(2000t100t2)R(t) = (1000 - t^2)(2000t - 100t^2)

We now differentiate R(t)R(t) with respect to tt. First, let's simplify the expression for R(t)R(t) by expanding the product:

R(t)=(1000×2000t)(1000×100t2)(t2×2000t)+(t2×100t2)R(t) = (1000 \times 2000t) - (1000 \times 100t^2) - (t^2 \times 2000t) + (t^2 \times 100t^2) R(t)=2,000,000t100,000t22000t3+100t4R(t) = 2,000,000t - 100,000t^2 - 2000t^3 + 100t^4

Now, differentiate R(t)R(t) with respect to tt:

R(t)=2,000,000200,000t6000t2+400t3R'(t) = 2,000,000 - 200,000t - 6000t^2 + 400t^3

Next, substitute t=5t = 5 months into the derivative:

R(5)=2,000,000200,000(5)6000(52)+400(53)R'(5) = 2,000,000 - 200,000(5) - 6000(5^2) + 400(5^3) R(5)=2,000,0001,000,0006000(25)+400(125)R'(5) = 2,000,000 - 1,000,000 - 6000(25) + 400(125) R(5)=2,000,0001,000,000150,000+50,000R'(5) = 2,000,000 - 1,000,000 - 150,000 + 50,000 R(5)=900,000 dollars per monthR'(5) = 900,000 \text{ dollars per month}

4. Interpretation

When t=5t = 5 months, the revenue is increasing at a rate of 900,000900,000 dollars per month.


Would you like more details on any of these calculations or explanations? Here are some related questions for further practice:

  1. What would the rate of change of sales be at t=10t = 10 months?
  2. How does the price change after 6 months? Calculate p(6)p'(6).
  3. Find the revenue at t=5t = 5 months (without differentiating).
  4. If the rate of change of price is negative, what does this mean in terms of business strategy?
  5. How does the rate of change of revenue behave as tt increases?

Tip: Always check whether the price and sales functions reflect real-world behavior over time, especially if long-term predictions are needed.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Revenue Functions
Rate of Change

Formulas

q(t) = 2,000t - 100t^2
p(t) = 1,000 - t^2
R(t) = p(t) * q(t)

Theorems

Product Rule of Differentiation
Chain Rule

Suitable Grade Level

Grades 11-12