Problem 45 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

OPTIMAL SELLING PRICE A retailer can obtain digital cameras from the manufacturer at a cost of \(\$ 150\) apiece. The retailer has been selling the cameras at the price of \(\$ 340\) apiece, and at this price, consumers have been buying 40 cameras a month. The retailer is planning to lower the price to stimulate sales and estimates that for each \(\$ 5\) reduction in the price, 10 more cameras will be sold each month. Express the retailer's monthly profit from the sale of the cameras as a function of the selling price. Draw the graph, and estimate the optimal selling price.

Step-by-Step Solution

Answer

The optimal selling price to maximize profit is \(\$330\).

Step 1 - Define Variables

Let the selling price be denoted by the variable \(p\). The cost price of each camera is \(\$150\). The initial selling price is \(\$340\), and initially, 40 cameras were sold.

Step 2 - Relation Between Price Reduction and Cameras Sold

For each \(\$5\) reduction in price, 10 more cameras are sold. If \(p\) is changed by \(x=\$5\), the new price can be expressed as \(340-5x\). And the new quantity sold is \(40 + 10x\).

Step 3 - Determine Revenue Function

Revenue is defined as the selling price times the quantity sold. So the revenue function \(R(x)\) in terms of \(x\) is \[ R(x) = p \times q = (340-5x) \times (40+10x). \] Substitute and simplify to get \[ R(x) = 13600 + 1700x - 50x^2. \]

Step 4 - Determine Cost Function

The cost is simply the cost price times the quantity sold. The cost function \(C(x)\) is given by \[ C(x) = 150 \times (40+10x) = 6000 + 1500x. \]

Step 5 - Determine Profit Function

Profit is revenue minus cost. The profit function \(P(x)\) is \[ P(x) = R(x) - C(x). \] Substituting the revenue and cost functions, \[ P(x) = (13600 + 1700x - 50x^2) - (6000 + 1500x) = 7600 + 200x - 50x^2. \]

Step 6 - Find the Vertex to Maximize Profit

To maximize profit, find the vertex of the parabola given by \[ P(x) = -50x^2 + 200x + 7600. \] The vertex form of a quadratic function \(ax^2 + bx + c\) is given by \[ x = \frac{-b}{2a}. \] Substituting \(a = -50\) and \(b = 200\), \[ x = \frac{-200}{2(-50)} = 2. \] Replace \(x\) in the price function gives \[ p = 340 - 5x = 340 - 5(2) = 330. \]

Step 7 - Conclusion

The optimal selling price is \(\$330\) to maximize monthly profit.

Key Concepts

Profit Function

The profit function is a crucial concept in determining the optimal selling price. Profit is simply the difference between revenue and cost. To express this mathematically:
\[ P(x) = R(x) - C(x) \] where \( R(x) \) is the revenue function and \( C(x) \) is the cost function.
In this exercise, we derived the profit function as:
\[ P(x) = 7600 + 200x - 50x^2 \] This quadratic equation helps us understand how the profit changes with the quantity sold.

Revenue Function

The revenue function represents the total money earned from sales before any costs are deducted. It is calculated as:
\[ R(x) = p \times q \] where \( p \) is the selling price and \( q \) is the quantity sold.
In the exercise, the revenue function was developed as:
\[ R(x) = (340 - 5x)(40 + 10x) \] After expanding and simplifying, it became:
\[ R(x) = 13600 + 1700x - 50x^2 \] This quadratic function helps us visualize how revenue changes with different selling prices.

Cost Function

The cost function calculates the total expense incurred in producing the goods sold. It is given by the product of the cost price per unit and the quantity sold:
\[ C(x) = c \times q \] where \( c \) is the manufacturing cost per unit.
For the given exercise, the cost function was:
\[ C(x) = 150 \times (40 + 10x) \] Simplifying this, we get:
\[ C(x) = 6000 + 1500x \] This linear function shows how costs increase with the number of units sold.

Quadratic Function

Quadratic functions are polynomial functions of degree two and are represented in the form:
\[ ax^2 + bx + c \] where \( a, b, \) and \( c \) are constants.
The profit function derived in the exercise:
\[ P(x) = 7600 + 200x - 50x^2 \] is an example of a quadratic function. Quadratic functions form parabolas when graphed, and their direction (opening upwards or downwards) is determined by the coefficient \( a \).
Here, \( a = -50 \), indicating the parabola opens downwards, helping us determine the maximum profit point.

Vertex of a Parabola

The vertex of a parabola represents the maximum or minimum point of a quadratic function. The vertex form of a parabola given by:
\[ ax^2 + bx + c \] is found using:
\[ x = \frac{-b}{2a} \]
In the exercise, we found the vertex of the profit function:
\[ P(x) = -50x^2 + 200x + 7600 \]
Substituting \( a = -50 \) and \( b = 200 \), we calculate:
\[ x = \frac{-200}{2(-50)} = 2 \]
This value of \( x \) helps us determine the optimal selling price when substituted back into the relationship between price and quantity, leading us to conclude the optimal price as \( \$330 \).