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Chapter 12: Problem 11

Assume that it costs Apple approximately $$C(x)=22,500+100 x+0.01 x^{2}$$ dollars to manufacture \(x 30\) -gigabyte video iPods in a day. How many iPods should be manufactured in order to minimize average cost? What is the resulting average cost of an iPod? (Give your answer to the nearest dollar.)

Short Answer

Expert verified
To minimize the average cost, Apple should manufacture approximately 2182 iPods per day, which results in an average cost of about $141 per iPod.

Step by step solution

01

Find the average cost function

Calculate the average cost function, denoted \(A(x)\), by dividing the cost function (\(C(x)\)) by the number of iPods (\(x\)): \[A(x) = \frac{C(x)}{x} = \frac{22500 + 100x + 0.01x^2}{x}\]
02

Calculate the derivative of the average cost function

To find the minimum average cost, we need to find the critical points of the average cost function. First, find the derivative of the average cost function, \(A(x)\), with respect to \(x\): \[A'(x) = \frac{d}{dx}\left(\frac{22500 + 100x + 0.01x^2}{x}\right)\] To differentiate, we can use the quotient rule, which states that \[\left(\frac{u}{v}\right)' = \frac{uv' - vu'}{v^2}\], where \(u = 22500 + 100x + 0.01x^2\) and \(v = x\). So we have: \[u'(x) = \frac{d}{dx}(22500 + 100x + 0.01x^2) = 100 + 0.02x\] \[v'(x) = \frac{d}{dx}(x) = 1\] Now, we can find the derivative using the quotient rule: \[A'(x) = \frac{x(100 + 0.02x) - (22500 + 100x + 0.01x^2)}{x^2}\] Simplify the derivative: \[A'(x) = \frac{0.02x^2 - 100x - 22500}{x^2}\]
03

Find critical points

Now, set the derivative equal to zero to find the critical points: \[0 = \frac{0.02x^2 - 100x - 22500}{x^2}\] Since the denominator cannot be zero, we only focus on the numerator: \[0 = 0.02x^2 - 100x - 22500\]
04

Solve for x

To solve for x, we can use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In this case, \(a = 0.02\), \(b = -100\), and \(c = -22500\). Plugging these values into the equation, we get: \[x = \frac{100 \pm \sqrt{(-100)^2 - 4(0.02)(-22500)}}{2(0.02)}\] Solving for x, we find two possible solutions: \(x \approx 2182.39\) and \(x \approx -206.39\). Since it is not possible to produce a negative number of iPods, we ignore the negative solution and consider \(x = 2182.39\).
05

Calculate the minimum average cost

Now that we have the number of iPods to minimize the average cost, we can plug this value back into the average cost function, \(A(x)\), to find the minimum average cost: \[A(2182.39) = \frac{22500 + 100(2182.39) + 0.01(2182.39)^2}{2182.39}\] Evaluating the expression gives us: \[A(2182.39) \approx 141.19\] Therefore, to minimize the average cost, Apple should manufacture approximately 2182 iPods per day, resulting in an average cost of around $141 per iPod.

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Key Concepts

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

Understanding the Cost Function
At the heart of minimizing average cost is the cost function, a mathematical representation that helps businesses calculate the total cost of producing a certain number of goods. In our example, Apple's cost function for manufacturing 30-gigabyte video iPods is given by:

\[ C(x) = 22,500 + 100x + 0.01x^2 \]
This equation consists of fixed costs (\(22,500\)), which are the costs that do not change regardless of the number of units produced, and variable costs (\(100x + 0.01x^2\)), which change with the level of output.

To understand how to minimize the average cost, imagine that each iPod's share of the fixed cost decreases as more units are produced, hence lowering the average cost up to a certain point. The variable costs, however, increase with each additional unit. The cost function, therefore, embodies the balance between these two types of costs.
Utilizing Derivative Calculus
To find the optimal number of iPods that minimizes the average cost, we appeal to derivative calculus. The derivative, in this context, is a powerful mathematical tool that tells us the rate at which costs are changing for each additional unit produced.

By calculating the derivative of the average cost function, we can identify the production level where the rate of change is zero鈥攊ndicating neither an increase nor decrease in average cost, which suggests a minimum or maximum point. The mathematical operation used here involves differentiating a fraction, which requires the quotient rule.

Once we obtain the derivative function \(A'(x)\), it gives us a formula that describes how the average cost changes with each additional iPod made. If the derivative is positive, the average cost is increasing; if it's negative, the average cost is decreasing. When the derivative is zero, we've likely found a turning point, which leads us to our next concept: the critical points.
Identifying Critical Points
In derivative calculus, critical points are where the function's derivative is zero or undefined. These points are crucial as they can indicate where the function achieves its maximum or minimum values鈥攊n our case, where the average cost is minimized.

After deriving the average cost function, we set the derivative equal to zero. This provides an equation that, when solved, yields potential critical points. However, not all critical points correspond with a minimum average cost. We must consider only realistic values鈥攊n this problem, negative production numbers are not possible, so we discard any negative solutions.

Critical points help inform businesses when they are operating at their most efficient in terms of cost management. While finding them involves complex calculus, the payoff is a clearer understanding of cost dynamics and optimal production levels. This analysis can be the difference between profit and loss in a competitive market.

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Most popular questions from this chapter

The approximate value of subprime (normally classified as risky) mortgage debt outstanding in the United States can be approximated by $$A(t)=\frac{1,350}{1+4.2(1.7)^{-t}} \text { billion dollars } \quad(0 \leq t \leq 8)$$ \(t\) years after the start of \(2000 .{ }^{38}\) Graph the function as well as its first and second derivatives. Determine, to the nearest whole number, the values of \(t\) for which the graph of \(A\) is concave up and concave down, and the \(t\) -coordinate of any points of inflection. What does the point of inflection tell you about subprime mortgages?

Rewrite the statements and questions in mathematical notation. The population \(P\) is currently 10,000 and growing at a rate of 1,000 per year.

Sketch the graph of the given function, indicating (a) \(x\) - and \(y\) -intercepts, (b) extrema, (c) points of inflection, \((d)\) behavior near points where the function is not defined, and (e) behavior at infinity. Where indicated, technology should be used to approximate the intercepts, coordinates of extrema, and/or points of inflection to one decimal place. Check your sketch using technology. \(g(t)=e^{-t^{2}}\)

Refer back to the model in the preceding exercise. The average number of computers per household in your town is \(0.5\) and is increasing at a rate of \(0.02\) computers per household per year. What is the average household income in your town, and how fast is it increasing? (Round your answers to the nearest \(\$ 10\) ).

Assume that it costs Microsoft approximately $$C(x)=14,400+550 x+0.01 x^{2}$$ dollars to manufacture \(x\) Xbox 360 s in a day. \(^{51}\) Obtain the average cost function, sketch its graph, and analyze the graph's important features. Interpret each feature in terms of Xboxes. HINT [Recall that the average cost function is \(\bar{C}(x)=C(x) / x\). \(]\)
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