Question

The total cost of producing and selling \(x\) units of Xbars per month is \(C(x)=100+3.002 x-0.0001 x^{2} .\) If the production level is 1600 units per month, find the average cost, \(C(x) / x\), of each unit and the marginal cost.

Short answer

Average cost is 2.9045; marginal cost is 2.682.

Step-by-step solution

Substitute into the Cost Function

To find the average cost, we first need to substitute the production level of 1600 units into the cost function. The cost function is given by:\[ C(x) = 100 + 3.002x - 0.0001x^2 \]Substitute \( x = 1600 \):\[ C(1600) = 100 + 3.002(1600) - 0.0001(1600)^2 \]

Compute the Total Cost

Calculate the total cost using the substituted value:\[ C(1600) = 100 + 3.002(1600) - 0.0001(2560000) \]\[ C(1600) = 100 + 4803.2 - 256 \]\[ C(1600) = 4647.2 \]

Calculate the Average Cost

The average cost per unit is found by dividing the total cost by the number of units produced. Thus, the average cost is:\[ \text{Average Cost} = \frac{C(1600)}{1600} \]\[ \text{Average Cost} = \frac{4647.2}{1600} \]\[ \text{Average Cost} = 2.9045 \]

Find the Marginal Cost Function

The marginal cost is the derivative of the cost function \( C(x) \). Differentiate the cost function with respect to \( x \):\[ C'(x) = \frac{d}{dx}(100 + 3.002x - 0.0001x^2) \]\[ C'(x) = 3.002 - 0.0002x \]

Evaluate the Marginal Cost at 1600 Units

Substitute \( x = 1600 \) into the marginal cost function to find the marginal cost:\[ C'(1600) = 3.002 - 0.0002(1600) \]\[ C'(1600) = 3.002 - 0.32 \]\[ C'(1600) = 2.682 \]

Key concepts

Cost Function

A cost function is an equation that represents the total cost associated with producing a certain number of goods. In this exercise, the cost function is given by:\[C(x) = 100 + 3.002x - 0.0001x^2\]The cost function is critical in economics and business, as it helps us understand how production and costs change with different levels of output. It typically includes fixed costs, variable costs per unit, and sometimes quadratic terms that model more complex cost behavior.Here, the term 100 is the fixed cost, meaning it does not change with the number of units produced. The term 3.002x represents the variable cost, which increases linearly with each additional unit. Lastly, the term \(-0.0001x^2\) is a quadratic component. It can suggest diminishing returns or economies of scale, as the cost per unit could decrease at certain production levels.

Marginal Cost

Marginal cost refers to the additional cost incurred by producing one more unit of a product. It is a vital concept in economics as it helps companies determine the optimal level of production. The marginal cost formula is derived from the cost function by taking its derivative.Here, the derivative of the cost function, or the marginal cost function, is:\[C'(x) = 3.002 - 0.0002x\]By evaluating the marginal cost at a specific production level, such as 1600 units, we can determine the cost to produce just one more unit beyond current production. In our exercise, when substituting 1600 into the marginal cost function, we obtain:\[C'(1600) = 2.682\]This marginal cost figure suggests the cost increase associated with producing an additional unit when already manufacturing 1600 units.

Derivative Calculation

Calculating the derivative of a function is essential to find the marginal cost. A derivative provides the rate at which one quantity changes with respect to another. In this context, it shows us how the cost changes as the number of units produced changes.The formula for the derivative of a function provides the slope of the function at any point \(x\). For the cost function \(C(x) = 100 + 3.002x - 0.0001x^2\), the derivative calculation results in:\[C'(x) = 3.002 - 0.0002x\]Each component of the cost function contributes to this derivative:

• The derivative of the constant 100 is 0, as constants do not change.
• The linear term 3.002x becomes 3.002, representing a constant slope or rate of change.
• The quadratic term \(-0.0001x^2\) becomes \(-0.0002x\), indicating a variable change depends on \(x\).

Differentiation helps in optimizing costs by identifying points where the cost function behaves differently.

Economics in Mathematics

Economics often relies on mathematical models to make sense of complex systems. In this exercise, mathematics is employed to calculate costs accurately and to evaluate how production changes impact overall expenses. Mathematical functions, like the cost function presented, allow economists to model real-world phenomena:

• By using derivatives, we can assess instantaneous rates of change—that's the essence of understanding marginal costs.
• Average cost calculations help in comparing total costs with production levels to make budgeting decisions.

Incorporating mathematics into economics provides clarity and precision, enabling better strategy planning and resource allocation. Mastery of these mathematical techniques is crucial for anyone pursuing a career in economics or business.