Question

The demand \(x\) for a web camera is 30,000 units per month when the price is \(\$ 25\) and 40,000 units when the price is \(\$ 20 .\) The initial investment is \(\$ 275,000\) and the cost per unit is \(\$ 17 .\) Assume that the demand is a linear function of the price. Find the profit \(\underline{P}\) as a function of \(x\) and approximate the change in profit for a one-unit increase in sales when \(x=28,000\). Make a sketch showing \(d P\) and \(\Delta P\).

Short answer

The profit function is given by \(P=-x^2/2000 + 28x - 275,000\). The derivative of the profit function evaluated at \(x=28,000\) yields \(d P=0\), meaning the change in profit for a one-unit increase in sales at \(x=28,000\) is zero. The sketch would show \(d P\) as a straight line with a negative slope, and \(\Delta P\) as a point on this line at \(d P=0\) when \(x=28,000\).

Step-by-step solution

Find the demand function

Since the demand \(x\) is said to be a linear function of the price, we can create two equations based on the provided data: (1) \(\$ 25\) for 30,000 units and (2) \(\$ 20\) for 40,000 units. These can be written as (1) \(30,000=25m+b\) and (2) \(40,000=20m+b\). Solving this system of equations for \(m\) (the slope) and \(b\) (the y-intercept), we get \(m=-2000\) and \(b=80000\). So the demand function is \(x=-2000p+80000\).

Formulating the Profit function

The profit function \(P\) can be formulated using the price \(p\), cost per unit, initial investment and the demand \(x\). \(p\) can be solved from the demand function as \(p=(80000-x)/2000\). The profit function is given as \(P=px - (cost \ per \ unit)x - initial \ investment\). Substituting the value of \(p\) in profit function, we get \(P=(80000x - x^2)/2000 - 17x - 275,000\) which simplifies to \(P=-x^2/2000 + 28x - 275,000\).

Find the change in Profit

The derivative of the profit function (\(d P\)) shows how the profit changes with a change in \(x\). Differentiating \(P\) with respect to \(x\), we get \(d P= -x/1000 + 28\).

Evaluate \(d P\) at \(x=28,000\)

Substituting \(x=28,000\) into the derivative, we get the change in profit for a one-unit increase in sales at \(x=28,000\) to be \(dP=-28+28=0\)

Sketching \(d P\) and \(\Delta P\)

\(d P\) is a straight line with a negative slope with an x-intercept at \(x=28,000\). The change in profit (\(\Delta P\)) at \(x=28,000\) is just a point on the \(d P\) line with \(x=28,000\) at \(d P=0\)

Key concepts

Linear Demand Function

The concept of a linear demand function plays a crucial role in understanding how the demand for a product changes with its price. In this case, the demand for web cameras depends linearly on the price. A linear demand function can be represented as a straight line equation, linking quantity demanded (\(x\)) to price (\(p\)). Given the problem's specifics, we are told that when the price is \(\\(25\), the demand is 30,000 units, and at \(\\)20\), it's 40,000 units. By constructing a system of equations based on these data points, we quickly find both the slope \(m\), which indicates how steeply demand changes with price, and the y-intercept \(b\), which tells us the demand level when the price crosses the y-axis.

• First equation: \(30,000 = 25m + b\)
• Second equation: \(40,000 = 20m + b\)

Solving this system, we find \(m = -2000\) and \(b = 80,000\), giving us the demand function: \[ x = -2000p + 80,000 \] This means for every dollar increase in price, demand decreases by 2000 units. Understanding this simple but powerful calculation allows us to predict how demand is likely to change with price adjustments.

Derivative in Profit Calculation

Calculating the derivative of the profit function is a key step in understanding how profit responds to changes in sales volume. Derivatives help us capture the rate of change of a function, which, in this context, is the profit function denoted by \(P\).The original problem translates the complex relationship of cost, price, and sales into a concise profit function: \[ P = -\frac{x^2}{2000} + 28x - 275,000 \] where \(x\) represents the number of units sold.By differentiating this profit function with respect to \(x\), we get:\[ dP = -\frac{x}{1000} + 28 \] This derivative reflects how profit \(P\) changes with each unit increase in sales. The derivative is linear itself, indicating a straightforward relationship where each sold unit affects profit predictably based on the current level of \(x\). This calculation is not only vital in maximizing profits but also in strategic pricing and sales decisions.

Change in Profit

To understand the change in profit, we employ both the derivative of the profit function and evaluate it at a specific value of \(x\). The formula for the change in profit \[ dP = -\frac{x}{1000} + 28 \] reveals that the change in profit is influenced directly by the number of units sold, \(x\). At a specific sales point, such as when \(x = 28,000\), substituting into the equation gives:\[ dP = -\frac{28,000}{1000} + 28 = 0 \]This result shows that at this point, increasing sales by one unit will not alter the profit. The balance of costs and revenue at this sales level creates a situation where a small increase in sales neither increases nor decreases profitability immediately. Understanding change in profit helps businesses determine the effectiveness of producing additional units and informs pricing and production strategies.

Cost Analysis

Analyzing costs is essential in understanding profits. Costs involve both initial and variable components. In the example, the initial investment is \(\\(275,000\), while the cost per web camera unit stands at \(\\)17\).The cost function becomes crucial when calculating the overall profit because profit is calculated as revenue minus costs. Thus, understanding both the initial (fixed) and variable costs is essential:

• Initial Investment: The fixed cost that does not change with the number of units, \(\\(275,000\).
• Variable Cost: The cost that varies with each unit produced, \(\\)17\) per unit.

To incorporate these into the profit calculation, you begin by calculating total variable costs based on production level \(x\): \[ \text{Total Variable Cost} = 17x \] The profit function combines these elements, subtracting both the fixed and total variable costs from total revenue:\[ P = ([80000x - x^2]/2000) - 17x - 275,000 \]Through cost analysis, businesses gain insight into necessary adjustments to price and production volume for achieving optimal profitability.