Question

The cost \(C\) (in dollars) of producing \(x\) units of a product is given by \(C=3.6 \sqrt{x}+500\) (a) Find the additional cost when the production increases from 9 to 10 units. (b) Find the marginal cost when \(x=9\). (c) Compare the results of parts (a) and (b).

Short answer

The additional cost is the difference between the cost for producing 10 units and the cost for producing 9 units, and can be computed directly from the function. The marginal cost is the derivative of the cost function evaluated at 9, and represents the rate of change of the cost at 9 units. The two values should be close since the marginal cost is an approximation of the actual cost difference for small changes in production.

Step-by-step solution

Compute the Difference in Cost

To find the additional cost when the production increases from 9 to 10 units, plug these values into the cost function. The costs are \(C(9) = 3.6 \sqrt{9} + 500\) and \(C(10) = 3.6 \sqrt{10} + 500\). Calculate these values and find the difference: \(C(10) - C(9)\)

Compute the Marginal Cost

The marginal cost is the derivative of the cost function. The given function is \(C(x) = 3.6 \sqrt{x} + 500\). Write \(\sqrt{x}\) as \(x^{1/2}\) and find the derivative, \(C'(x)\), using power rule. Then, find \(C'(9)\), the marginal cost when \(x = 9\)

Compare the Results

Compare the additional cost from Step 1 and the marginal cost from Step 2. The marginal cost represents the theoretical cost of producing one more unit, while the additional cost is the actual cost difference for producing one more unit.

Key concepts

Cost Function

At its core, a cost function is a mathematical formula that a business uses to calculate the expenses associated with producing a given number of goods. It represents the total production cost as a function of quantity. Take, for example, the cost function provided in our exercise:
\(C = 3.6 \sqrt{x} + 500\),
where \(C\) represents the total cost in dollars, and \(x\) is the number of units produced. The constant term, in this case, 500, might include fixed costs such as rent and utilities, which do not change with the level of production. The variable component, \(3.6 \sqrt{x}\), likely represents variable costs that increase with more units produced, such as materials and labor. Understanding this function is crucial for businesses because it helps them determine pricing strategies, break-even points, and profitability thresholds.

Derivative of a Function

The derivative of a function at any point gives the rate at which the function value is changing at that point with respect to its input variable. In economic terms, the derivative of the cost function represents the marginal cost, which is the cost of producing one additional unit. This concept is particularly useful for production cost analysis.
To find this derivative, mathematicians utilize calculus tools, such as the power rule. For instance, in the function provided, first we express \(\sqrt{x}\) as \(x^{1/2}\). The derivative, noted as \(C'(x)\), is then obtained by multiplying the exponent by the coefficient and subtracting one from the exponent, resulting in the derivative function. In our example, the marginal cost at a production level of \(x = 9\) units is found by computing \(C'(9)\).

Production Cost Analysis

Production cost analysis involves examining the various costs associated with the production of goods or services. It's a crucial aspect of decision-making for any business. In this context, understanding both the total cost and the marginal cost is essential. The total cost provides the overall expenditure for creating a certain number of goods, while the marginal cost, as discussed earlier, is the cost of producing an additional unit.
Businesses use these analyses to make decisions about scaling production, setting prices, and optimizing efficiency. By comparing the actual additional cost of increasing production with the theoretical marginal cost calculated using calculus, companies can assess how well their actual costs align with the theoretical models, allowing for more precise and efficient production strategies.