Question

The monthly revenue \(R\) achieved by selling \(x\) wristwatches is \(R(x)=75 x-0.2 x^{2} .\) The monthly cost \(C\) of selling \(x\) wristwatches is $$ C(x)=32 x+1750 $$ (a) How many wristwatches must the firm sell to maximize revenue? What is the maximum revenue? (b) Profit is given as \(P(x)=R(x)-C(x)\). What is the profit function? (c) How many wristwatches must the firm sell to maximize profit? What is the maximum profit? (d) Provide a reasonable explanation as to why the answers found in parts (a) and (c) differ. Explain why a quadratic function is a reasonable model for revenue.

Short answer

Sell 187.5 wristwatches for maximum revenue; the maximum revenue is 7031.25. The profit function is \( P(x) = 43x - 0.2x^2 - 1750 \). Sell 107.5 wristwatches for maximum profit; the maximum profit is 561.25.

Step-by-step solution

Determine the Revenue Function

The monthly revenue function is given by \( R(x) = 75x - 0.2x^2 \).

Find the Vertex of the Revenue Function

To find the number of wristwatches that maximize revenue, we need to find the vertex of the quadratic function. The vertex form for a quadratic equation \( ax^2 + bx + c \) is given by \( x = -\frac{b}{2a} \). Here, \( a = -0.2 \) and \( b = 75 \).\[ x = -\frac{75}{2(-0.2)} = \frac{75}{0.4} = 187.5 \]So, the firm must sell 187.5 wristwatches to maximize revenue.

Calculate Maximum Revenue

Substitute \( x = 187.5 \) into the revenue function to find the maximum revenue.\[ R(187.5) = 75(187.5) - 0.2(187.5)^2 \]\[ R(187.5) = 14062.5 - 0.2(35156.25) = 14062.5 - 7031.25 = 7031.25 \]The maximum revenue is 7031.25.

Determine the Profit Function

The profit function is given by the revenue minus the cost: \( P(x) = R(x) - C(x) \). Substituting the given functions, we get:\[ P(x) = (75x - 0.2x^2) - (32x + 1750) \]\[ P(x) = 75x - 0.2x^2 - 32x - 1750 \]\[ P(x) = 43x - 0.2x^2 - 1750 \]

Find the Vertex of the Profit Function

To find the number of wristwatches that maximize profit, we need to find the vertex of the profit function. For the quadratic equation \( P(x) = -0.2x^2 + 43x - 1750 \), the vertex is given by \( x = -\frac{b}{2a} \).Here, \( a = -0.2 \) and \( b = 43 \).\[ x = -\frac{43}{2(-0.2)} = \frac{43}{0.4} = 107.5 \]So, the firm must sell 107.5 wristwatches to maximize profit.

Calculate Maximum Profit

Substitute \( x = 107.5 \) into the profit function to find the maximum profit.\[ P(107.5) = 43(107.5) - 0.2(107.5)^2 - 1750 \]\[ P(107.5) = 4622.5 - 0.2(11556.25) - 1750 \]\[ P(107.5) = 4622.5 - 2311.25 - 1750 = 561.25 \]The maximum profit is 561.25.

Explanation of Differences in Maximum Revenue and Profit

The difference in the number of wristwatches sold to maximize revenue and profit results from the cost function. While revenue only considers the earnings from sales, profit accounts for both earnings and costs. As costs increase with the number of wristwatches sold, the optimal number of wristwatches for maximizing profit is lower than that for maximizing revenue.

Reasonableness of the Quadratic Revenue Model

A quadratic model for revenue is reasonable because it captures the effect of diminishing returns. Initially, increasing the number of wristwatches sold increases revenue, but after a certain point, additional sales can lead to negative effects such as lower prices or increased production costs, thus reducing the overall revenue.

Key concepts

Revenue Function

In this problem, the revenue function is given by the equation \( R(x) = 75x - 0.2x^2 \). This function helps us understand how revenue changes with the number of wristwatches sold (represented by \( x \)). Generally, revenue functions like these are quadratic, implying they have a parabolic shape.
The revenue function here is maximized where the number of watches sold allows for the highest revenue before diminishing returns kick in. By finding the vertex of this parabola, we can determine the point at which revenue is maximized.
The general form of a quadratic function is \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. In our case, \( a = -0.2 \), \( b = 75 \), and \( c = 0 \). The negative coefficient of \( x^2 \) indicates the parabola opens downwards, meaning it has a maximum point.

Profit Function

The profit function is crucial because it tells us how much the firm actually earns after subtracting costs from revenue. The formula is \( P(x) = R(x) - C(x) \).
For this problem:
The revenue function is \( R(x) = 75x - 0.2x^2 \), and the cost function is \( C(x) = 32x + 1750 \).
By subtracting the cost function from the revenue function, the profit function is:
\( P(x) = (75x - 0.2x^2) - (32x + 1750) \)
\( P(x) = 43x - 0.2x^2 - 1750 \)
This equation shows the relationship between the number of watches sold and the profit earned.
To find the number of watches that maximizes profit, we need to find the vertex of this quadratic function, mirroring the approach for maximizing revenue.

Vertex of a Quadratic Function

The vertex of a quadratic function \( ax^2 + bx + c \) is found using the formula \( x = -\frac{b}{2a} \). The vertex represents the highest or lowest point on the parabola. In the context of revenue and profit functions, this point shows where revenue or profit is maximized or minimized.
For the revenue function \( R(x) = 75x - 0.2x^2 \), the vertex is at:
\[ x = -\frac{75}{2(-0.2)} = \frac{75}{0.4} = 187.5 \]
This tells us that selling 187.5 wristwatches will yield the maximum revenue.
For the profit function \( P(x) = 43x - 0.2x^2 - 1750 \), the vertex is at:
\[ x = -\frac{43}{2(-0.2)} = \frac{43}{0.4} = 107.5 \]
This tells us the firm should sell 107.5 wristwatches to maximize profit. The difference in number is because profit considers costs, unlike revenue.

Quadratic Revenue Model

Using a quadratic model for revenue is practical because it captures real-world behavior, where increasing sales initially boosts revenue, but beyond a certain point, additional sales can decrease overall revenue.
This reflects the principle of diminishing returns: initially, increasing production leads to more revenue, but eventually, factors such as market saturation or increased costs lower the overall revenue.
In the given revenue function \( R(x) = 75x - 0.2x^2 \), at first, each additional wristwatch sold increases revenue, but as \( x \) grows larger, the term \( -0.2x^2 \) increases faster, offsetting the gains, and thus eventually reducing total revenue.
Understanding this concept helps businesses optimize their output to maximize revenue and profit effectively.