Math Problem Statement

An art gallery offers 30 prints by a famous artist. If each print in the limited edition is priced at p dollars, it is expected that q=580-2p prints will be sold.

a. Find the elasticity of demand. Determine the values of p for which the demand is elastic, inelastic, and of unit elasticity. b. If you were the owner of the gallery, what price would you charge for each print? Explain the reasoning behind your decision. Gr

Solution

Part (a): Elasticity of Demand

The formula for the elasticity of demand E(p)E(p) is given by:

E(p)=pqdqdpE(p) = \frac{p}{q} \cdot \frac{dq}{dp}

Where:

  • pp is the price,
  • q=5802pq = 580 - 2p is the quantity demanded as a function of pp.

First, calculate dqdp\frac{dq}{dp}:

q=5802pq = 580 - 2p dqdp=2\frac{dq}{dp} = -2

Now, using the elasticity formula:

E(p)=pq(2)E(p) = \frac{p}{q} \cdot (-2)

Substitute q=5802pq = 580 - 2p:

E(p)=p5802p(2)E(p) = \frac{p}{580 - 2p} \cdot (-2)

This simplifies to:

E(p)=2p5802pE(p) = \frac{-2p}{580 - 2p}

Elastic, Inelastic, and Unit Elastic:

  • Elastic demand occurs when E(p)>1E(p) > 1.
  • Inelastic demand occurs when E(p)<1E(p) < 1.
  • Unit elasticity occurs when E(p)=1E(p) = 1.

To find the price at which demand is unit elastic (E(p)=1E(p) = -1):

2p5802p=1\frac{-2p}{580 - 2p} = -1

Solve for pp:

2p=5802p2p = 580 - 2p 4p=5804p = 580 p=145p = 145

  • Unit Elastic at p=145p = 145.

For elastic demand (E(p)<1E(p) < -1):

2p5802p<1\frac{-2p}{580 - 2p} < -1

Simplifying:

2p>5802p2p > 580 - 2p 4p>5804p > 580 p>145p > 145

For inelastic demand (E(p)>1E(p) > -1):

2p5802p>1\frac{-2p}{580 - 2p} > -1

Simplifying:

2p<5802p2p < 580 - 2p 4p<5804p < 580 p<145p < 145

Conclusion for Part (a):

  • Demand is elastic when p>145p > 145.
  • Demand is inelastic when p<145p < 145.
  • Demand is of unit elasticity when p=145p = 145.

Part (b): Optimal Price to Charge

As a gallery owner, you would want to maximize revenue. Revenue R(p)R(p) is given by:

R(p)=pq=p(5802p)R(p) = p \cdot q = p(580 - 2p)

Expanding this:

R(p)=580p2p2R(p) = 580p - 2p^2

To maximize revenue, we take the derivative of R(p)R(p) with respect to pp and set it equal to zero:

dRdp=5804p\frac{dR}{dp} = 580 - 4p

Set dRdp=0\frac{dR}{dp} = 0 to find the critical point:

5804p=0580 - 4p = 0 4p=5804p = 580 p=145p = 145

Since p=145p = 145 is also the price at which demand is unit elastic, this is the optimal price to charge for maximizing revenue.

Conclusion for Part (b):

You should charge $145 per print. At this price:

  • Revenue is maximized.
  • Demand is unit elastic, meaning a 1% change in price will result in a 1% change in quantity demanded, which is the best balance between price and sales.

Do you want further details on elasticity, or have any specific questions?

Here are 5 related questions to explore:

  1. How is revenue affected when demand is elastic or inelastic?
  2. How does marginal cost influence pricing strategies in this scenario?
  3. What would happen if the prints were priced at $200?
  4. How does the concept of price discrimination apply to art galleries?
  5. Can other factors, such as artist popularity, affect demand elasticity in this case?

Tip: When optimizing prices, always consider both demand elasticity and external factors like consumer preferences and market trends.

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Math Problem Analysis

Mathematical Concepts

Elasticity of Demand
Revenue Maximization
Calculus

Formulas

E(p) = (p/q) * (dq/dp)
q = 580 - 2p
Revenue R(p) = p * q = p(580 - 2p)

Theorems

Elasticity of Demand Theorem
Revenue Maximization using Derivatives

Suitable Grade Level

Undergraduate Economics/Business Calculus