Math Problem Statement

​(A​)

Find

StartFraction dx Over dp EndFractiondxdp.

StartFraction dx Over dp EndFractiondxdpequals=StartFraction negative 300 Over left parenthesis 0.3 p plus 1 right parenthesis squared EndFraction−300(0.3p+1)2

Part 2

​(B​)

Find the demand and the instantaneous rate of change of demand with respect to price when the price is

​$330.

Write a brief interpretation of these results.

The demand is

xequals=100100

when the price is

​$330.

Part 3

The instantaneous rate of change of demand with respect to price is

negative 3−3

when the price is

​$330.

Part 4

Write a brief interpretation of these results.

At a price level of

​$3030​,

the demand is

100100

DVD players per week and demand is

decreasing

at the rate of

33

DVD players per dollar

.

Part 5

​(C​)

Use the results from part

​(B​)

to estimate the demand if the price is increased to

​$331.

The demand is approximately

enter your response here

if the price is increased to

​$331.

Solution

Let's go step by step to understand and solve this problem.

Part 1 (A) — Finding the derivative of demand with respect to price:

We are given that:

dxdp=300(0.3p+1)2\frac{dx}{dp} = \frac{-300}{(0.3p + 1)^2}

This represents the rate of change of demand (xx) with respect to price (pp).

Part 2 (B) — Find the demand and the instantaneous rate of change of demand with respect to price when the price is $330:

  • Demand when the price is $330: The demand xx is given as 100 when the price pp is $330. This is directly stated in the problem.

  • Instantaneous rate of change of demand with respect to price: We can calculate the rate of change of demand using the formula we already have:

dxdp=300(0.3p+1)2\frac{dx}{dp} = \frac{-300}{(0.3p + 1)^2}

Substitute p=330p = 330:

dxdp=300(0.3(330)+1)2\frac{dx}{dp} = \frac{-300}{(0.3(330) + 1)^2}

Simplify:

dxdp=300(99+1)2=3001002=30010000=0.03\frac{dx}{dp} = \frac{-300}{(99 + 1)^2} = \frac{-300}{100^2} = \frac{-300}{10000} = -0.03

So, the instantaneous rate of change of demand with respect to price when the price is $330 is 0.03-0.03. This means the demand is decreasing by 0.03 units per dollar when the price is $330.

Part 3 (Interpretation of Results):

At a price of $330, the demand is 100 units (DVD players per week), and the demand is decreasing at a rate of 3 units per $1 (negative 0.03), which means that for every additional dollar increase in price, the demand decreases by 3 units.

Part 4 (C) — Estimate the demand when the price is increased to $331:

To estimate the demand when the price increases to $331, we can use the linear approximation based on the rate of change.

We know that:

  • The demand when the price is $330 is 100 units.
  • The rate of change of demand when the price is $330 is 3-3 units per dollar.

The approximate change in demand Δx\Delta x for a small increase in price (Δp=1\Delta p = 1) is given by:

ΔxdxdpΔp\Delta x \approx \frac{dx}{dp} \cdot \Delta p

Substitute the values:

Δx31=3\Delta x \approx -3 \cdot 1 = -3

Thus, the demand at p=331p = 331 is approximately:

x(331)=x(330)+Δx=1003=97x(331) = x(330) + \Delta x = 100 - 3 = 97

So, the estimated demand when the price is increased to $331 is 97 units.

Final Summary:

  • Demand at $330 = 100 units
  • Rate of change of demand at $330 = 3-3 units per dollar
  • Estimated demand at $331 = 97 units

Would you like any further details on any of these steps? Let me know if you need more clarification!


Further Questions for Exploration:

  1. How do you interpret the relationship between demand and price based on this model?
  2. How does the derivative dxdp\frac{dx}{dp} affect the demand curve for a product?
  3. What would happen if the rate of change of demand were positive instead of negative?
  4. How would the estimation of demand change if the price increase were larger than 1 dollar?
  5. How do changes in the rate of change dxdp\frac{dx}{dp} influence the long-term trend in demand?

Tip: When estimating demand changes, using the linear approximation formula gives a quick estimate but is most accurate for small changes in price!

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Rates of Change
Demand Function

Formulas

dx/dp = -300 / (0.3p + 1)^2
Linear Approximation: Δx ≈ (dx/dp) * Δp

Theorems

Derivative as Rate of Change
Linear Approximation

Suitable Grade Level

Grades 11-12, College-level Calculus