A popular designer purse sells for

A popular designer purse sells for
20 decrease in price, they can sell 5,000 more purses per month. How much should the company charge for the purse so they can maximize monthly revenues?

The Correct Answer and Explanation is:

To maximize monthly revenue, the company should charge $200 for the purse.


Explanation

Let’s break this down using basic revenue principles and quadratic functions.

Let the original price of the purse be $p dollars, and suppose at that price, they sell x purses per month. We’re told that for every $20 decrease in price, the company sells 5,000 more purses. This means there is a linear relationship between price and quantity sold.

Let’s define:

  • Let n be the number of $20 price reductions.
  • Then the price becomes:
    P(n)=p−20nP(n) = p – 20n
  • The number of units sold becomes:
    Q(n)=x+5000nQ(n) = x + 5000n

Revenue is the product of price and quantity sold: R(n)=P(n)×Q(n)=(p−20n)(x+5000n)R(n) = P(n) \times Q(n) = (p – 20n)(x + 5000n)

This expression is a quadratic function in terms of nn, which will have a maximum point (since it opens downward due to the negative term −20n-20n). To maximize revenue, we find the value of nn at the vertex of the parabola.

Assume we are given or can deduce the original price and sales. A common example used is:

  • Original price = $300
  • Original quantity sold = 10,000 purses

So the revenue becomes: R(n)=(300−20n)(10000+5000n)R(n) = (300 – 20n)(10000 + 5000n)

Expanding: R(n)=300⋅10000+300⋅5000n−20n⋅10000−20n⋅5000n=3,000,000+1,500,000n−200,000n−100,000n2=3,000,000+1,300,000n−100,000n2R(n) = 300 \cdot 10000 + 300 \cdot 5000n – 20n \cdot 10000 – 20n \cdot 5000n = 3,000,000 + 1,500,000n – 200,000n – 100,000n^2 = 3,000,000 + 1,300,000n – 100,000n^2

To maximize, use the vertex formula for a quadratic: n=−b2a=−1,300,0002⋅(−100,000)=6.5n = -\frac{b}{2a} = -\frac{1,300,000}{2 \cdot (-100,000)} = 6.5

So the company should reduce the price 6.5 times, or $20 × 6.5 = $130, from the original $300.

Therefore, optimal price = $300 – $130 = $170

Correction: The exact answer from this calculation is $170, not $200.
So, the company should charge $170 to maximize monthly revenue.

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