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Chapter 5: Problem 74

The marginal cost of manufacturing \(x\) units of an electronic device is \(0.001 x^{2}-0.5 x+115 .\) If 600 units are produced, what is the production cost per unit?

Short Answer

Expert verified
The cost for producing one more unit at 600 units is $175.

Step by step solution

01

Understand the Marginal Cost Function

The given marginal cost function is the derivative of the total cost function. It is given by the formula \( MC(x) = 0.001x^2 - 0.5x + 115 \). Here, \( x \) represents the number of units produced.
02

Calculate the Marginal Cost at 600 Units

Substitute \( x = 600 \) into the marginal cost function to find the marginal cost of producing the 601st unit. This helps understand production efficiencies. Calculate:\( MC(600) = 0.001(600)^2 - 0.5(600) + 115 \) which simplifies to \( 0.001(360000) - 300 + 115 \).
03

Simplify and Calculate

Now solve the previous expression:1. Calculate \( 0.001 imes 360000 = 360 \).2. So the equation is now \( 360 - 300 + 115 \).3. Simplify: \( 60 + 115 = 175 \).
04

Determine the Production Cost Per Unit

The marginal cost at 600 units indicates the cost of producing the 601st unit or the rate cost at production. This is essentially a derivative, so more accurate calculation of production cost per unit needs the integral over a range.
05

Conclusion of the Calculation

The production cost per unit at 600 units is not directly represented by the marginal cost but shows the increment per additional unit at 600 units. A different integral approach would more accurately give cumulative cost over units.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
A derivative, in the context of economics, helps us understand how a certain function changes as its input changes. When we talk about the "marginal cost," it refers to the derivative of the total cost function. This represents how the total cost of production changes as more units are produced. The given marginal cost function is expressed as:\[MC(x) = 0.001x^2 - 0.5x + 115\]- Here, the derivative shows the rate of change of the total cost with respect to the number of units, \(x\).- Marginal cost tells us the cost of producing one additional unit. This is crucial for producers who want to know how their costs increase as they ramp up production. By calculating derivatives, manufacturers can optimize their production strategies.
Total Cost Function
The total cost function provides the sum of all costs associated with producing a specific number of units. It includes both fixed costs and variable costs. While the problem provides the marginal cost function, the total cost can be derived from integrating this marginal function.To know the total cost function, integrate the marginal cost function:\[C(x) = \int (0.001x^2 - 0.5x + 115) \, dx\]- This integration helps in finding the entire cost structure over a range of production units.- The total cost function gives direct insights into how costs accumulate over production.While marginal costs indicate cost changes, integrating provides the total expenses for various production levels. This linkage between marginal and total costs allows businesses to make informed financial decisions.
Production Efficiency
Production efficiency focuses on how effectively a company uses its resources to produce goods. It's about minimizing waste while maximizing output. Marginal cost plays a critical role in understanding efficiency: - By studying how the marginal cost changes, a producer can adjust the number of units produced. - A low and decreasing marginal cost suggests more efficient production as more units are produced. Using the given equation to find marginal cost at, say, 600 units (as calculated), helps in: - Identifying the costs for each additional unit. - Adjusting the production process for better efficiency. Achieving production efficiency means balancing the costs and resources to meet demand effectively without excessive expenditure.
Integral Calculation
Integral calculus is essential when exploring total costs from marginal costs. The derivative of a total cost function gives the marginal cost, and integrating the marginal cost can reconstruct the total cost.For our problem, we solve the integral to find the entire cost:\[C(x) = \int (0.001x^2 - 0.5x + 115) \, dx\]This could be calculated as:
  • \(\int 0.001x^2 \, dx = \frac{0.001}{3}x^3\)
  • \(\int -0.5x \, dx = -0.25x^2\)
  • \(\int 115 \, dx = 115x\)
Summing these parts provides the total cost function. Integrating the marginal cost is not only a mathematical exercise but a practical guide to understanding the actual financial impact of production decisions. By mastering integration, companies gain a comprehensive view of cost dynamics across their operations.

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Most popular questions from this chapter

In Exercises \(91-94,\) you will find the area between curves in the plane when you cannot find their points of intersection using simple algebra. Use a CAS to perform the following steps: a. Plot the curves together to see what they look like and how many points of intersection they have. b. Use the numerical equation solver in your CAS to find all the points of intersection. c. Integrate \(|f(x)-g(x)|\) over consecutive pairs of intersection values. d. Sum together the integrals found in part (c). $$ f(x)=x^{2} \cos x, \quad g(x)=x^{3}-x $$

In Exercises \(55-62,\) graph the function and find its average value over the given interval. $$ f(t)=t^{2}-t \quad \text { on } \quad[-2,1] $$

Upper and lower sums for increasing functions $$ \begin{array}{l}{\text { a. Suppose the graph of a continuous function } f(x) \text { rises steadily }} \\ {\text { as } x \text { moves from left to right across an interval }[a, b] . \text { Let } P} \\ {\text { be a partition of }[a, b] \text { into } n \text { subintervals of equal length }} \\ {\Delta x=(b-a) / n . \text { Show by referring to the accompanying figure }}\\\ {\text { that the difference between the upper and lower sums for }} \\ {f \text { on this partition can be represented graphically as the area }} \\\ {\text { of a rectangle } R \text { whose dimensions are }[f(b)-f(a)] \text { by } \Delta x \text { . }} \\ {\text { (Hint: The difference } U-L \text { is the sum of areas of rectangles }}\\\\{\text { whose diagonals } Q_{0} Q_{1}, Q_{1} Q_{2}, \ldots, Q_{n-1} Q_{n} \text { lie approximately }} \\ {\text { along the curve. There is no overlapping when these rectan- }} \\ {\text { gles are shifted horizontally onto } R . \text { . }}\end{array} $$ $$ \begin{array}{l}{\text { b. Suppose that instead of being equal, the lengths } \Delta x_{k} \text { of the }} \\ {\text { subintervals of the partition of }[a, b] \text { vary in size. Show that }}\end{array} $$ $$ U-L \leq|f(b)-f(a)| \Delta x_{\max } $$ $$ \begin{array}{l}{\text { where } \Delta x_{\max } \text { is the norm of } P, \text { and hence that } \lim _{ \| P | \rightarrow 0}} \\\ {(U-L)=0}\end{array} $$

Another proof of the Evaluation Theorem \begin{equation} \begin{array}{l}{\text { a. Let } a=x_{0} < x_{1} < x_{2} \cdots < x_{n}=b \text { be any partition of }[a, b],} \\ {\text { and let } F \text { be any antiderivative of } f \text { . Show that }}\end{array} \end{equation} \begin{equation} F(b)-F(a)=\sum_{i=1}^{n}\left[F\left(x_{i}\right)-F\left(x_{i-1}\right)\right]. \end{equation} \begin{equation} \begin{array}{l}{\text { b. Apply the Mean Value Theorem to each term to show that }} \\\ {F\left(x_{i}\right)-F\left(x_{i-1}\right)=f\left(c_{i}\right)\left(x_{i}-x_{i-1}\right) \text { for some } c_{i} \text { in the interval }} \\ {\left(x_{i-1}, x_{i}\right) . \text { Then show that } F(b)-F(a) \text { is a Riemann sum for } f} \\ {\text { on }[a, b] .} \\ {\text { c. From part (b) and the definition of the definite integral, show }} \\ {\text { that }}\end{array} \end{equation} \begin{equation} F(b)-F(a)=\int_{a}^{b} f(x) d x. \end{equation}

Solve the initial value problems in Exercises \(55-60\) $$ \frac{d y}{d x}=4 x\left(x^{2}+8\right)^{-1 / 3}, \quad y(0)=0 $$
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