Problem 88
Question
The cost \(C\) of producing \(x\) units of a product is given by \(C=0.2 x^{2}+10 x+5,\) and the average cost per unit is given by $$\bar{C}=\frac{C}{x}=\frac{0.2 x^{2}+10 x+5}{x}, \quad x>0$$ Sketch the graph of the average cost function, and estimate the number of units that should be produced to minimize the average cost per unit.
Step-by-Step Solution
Verified Answer
You should produce 5 units to minimize the average cost.
1Step 1: Deriving the Average Cost Function
The average cost function \(\bar{C}=\frac{C}{x}=\frac{0.2 x^{2}+10 x+5}{x}\). The first step is to simplify it: \(\bar{C}=0.2 x + 10 + \frac{5}{x}\), where \(x > 0\).
2Step 2: Sketching the Average Cost Function
To sketch this graph, plot the function for a reasonable range of \(x\) values. Ideally, this graph should be downward-opening, since \(\bar{C}\) decreases as \(x\) increases.
3Step 3: Minimizing the Average Cost Function
The minimum average cost can be found by differentiating the average cost function with respect to \(x\) and solving for \(x\) when the derivative is zero. The derivative of \(\bar{C}=0.2 x + 10 + \frac{5}{x}\) is \(\bar{C}' = 0.2 - \frac{5}{x^2}\). Find the value of \(x\) when \(\bar{C}' = 0\): \(0 = 0.2 - \frac{5}{x^2}\), which gives \(x = \sqrt(\frac{5}{0.2}) = \sqrt{25} = 5\).
4Step 4: Verifying the Minimum
To confirm that \(x = 5\) gives a minimum (not a maximum), find the second derivative and verify that it is positive at this value. The second derivative of \(\bar{C}\) is \(\bar{C}'' = \frac{10}{x^3}\), and for \(x = 5\): \(\bar{C}'' = \frac{10}{125} > 0\).
Key Concepts
Cost Production FunctionDerivative OptimizationGraphing FunctionsCalculus in Economics
Cost Production Function
Understanding the cost production function is vital for economic decision-making within a business. It represents the total cost of producing a certain number of units of a product and is typically composed of fixed and variable costs. In our example, the cost production function is given by the equation
\[ C = 0.2x^2 + 10x + 5 \]
where \(x\) denotes the number of units produced, \(0.2x^2\) represents the variable costs that change with the production level, \(10x\) indicates a linear relationship for a different dimension of variable costs, and \(5\) is the fixed cost that remains constant regardless of the production volume.
Essentially, the shape and nature of this function reflect the interplay between economies of scale and the spreading of fixed costs over a larger number of units.
\[ C = 0.2x^2 + 10x + 5 \]
where \(x\) denotes the number of units produced, \(0.2x^2\) represents the variable costs that change with the production level, \(10x\) indicates a linear relationship for a different dimension of variable costs, and \(5\) is the fixed cost that remains constant regardless of the production volume.
Essentially, the shape and nature of this function reflect the interplay between economies of scale and the spreading of fixed costs over a larger number of units.
Derivative Optimization
The concept of derivative optimization comes into play when we aim to minimize or maximize a particular function. By taking the derivative of a function, we determine the rate at which it changes with respect to a given variable.
In economic contexts, like optimizing the average cost, we find the derivative of the average cost function \(\bar{C}\) to locate critical points, which are candidates for minimum or maximum values. Once the derivative is set to zero, solving for the variable will give the quantity that minimizes or maximizes the function. The derivative of our average cost function is
\[ \bar{C}' = 0.2 - \frac{5}{x^2} \]
By setting \(\bar{C}' = 0\), we find the production level that theoretically yields the lowest average cost. Further examination with the second derivative confirms whether we have found a minimum (second derivative is positive) or maximum (second derivative is negative).
In economic contexts, like optimizing the average cost, we find the derivative of the average cost function \(\bar{C}\) to locate critical points, which are candidates for minimum or maximum values. Once the derivative is set to zero, solving for the variable will give the quantity that minimizes or maximizes the function. The derivative of our average cost function is
\[ \bar{C}' = 0.2 - \frac{5}{x^2} \]
By setting \(\bar{C}' = 0\), we find the production level that theoretically yields the lowest average cost. Further examination with the second derivative confirms whether we have found a minimum (second derivative is positive) or maximum (second derivative is negative).
Graphing Functions
Graphing functions serves as a powerful visual tool in understanding and analyzing the behavior of functions. It allows us to see at a glance how a variable responds to changes in another related variable. When sketching the graph of the average cost function, for example, it is helpful to note that as \(x\) increases, the average cost \(\bar{C}\) generally decreases due to economies of scale, until reaching the minimum cost.
To properly sketch the average cost function,
\[ \bar{C} = 0.2x + 10 + \frac{5}{x} \]
one should select a range of \(x\) values, plot the corresponding \(\bar{C}\) values, and then connect these points smoothly. The graph typically reveals a U-shape, reflecting the initial decrease in average costs and subsequent increase if production is pushed beyond the efficient scale.
To properly sketch the average cost function,
\[ \bar{C} = 0.2x + 10 + \frac{5}{x} \]
one should select a range of \(x\) values, plot the corresponding \(\bar{C}\) values, and then connect these points smoothly. The graph typically reveals a U-shape, reflecting the initial decrease in average costs and subsequent increase if production is pushed beyond the efficient scale.
Calculus in Economics
Calculus plays a fundamental role in economics, helping to optimize outcomes such as profit maximization and cost minimization. It offers tools to analyze the effects of small changes in economic variables on overall outcomes.
In the given problem, calculus allows us to minimize the average cost of production by taking derivatives and finding critical points. This mathematical approach is essential in understanding not just theoretical models, but also in applying real-world business strategies that require the calculation of marginal costs and benefits. By setting the derivative of the average cost function to zero, calculus gives us the precise quantity of production that delivers the most cost-efficient result.
In the given problem, calculus allows us to minimize the average cost of production by taking derivatives and finding critical points. This mathematical approach is essential in understanding not just theoretical models, but also in applying real-world business strategies that require the calculation of marginal costs and benefits. By setting the derivative of the average cost function to zero, calculus gives us the precise quantity of production that delivers the most cost-efficient result.
Other exercises in this chapter
Problem 87
The ordering and transportation cost \(C\) (in thousands of dollars) for the components used in manufacturing a product is given by $$C=100\left(\frac{200}{x^{2
View solution Problem 87
Sketch the graph of the function by (a) applying the Leading Coefficient Test, (b) finding the zeros of the polynomial, (c) plotting sufficient solution points,
View solution Problem 88
Sketch the graph of the function by (a) applying the Leading Coefficient Test, (b) finding the zeros of the polynomial, (c) plotting sufficient solution points,
View solution Problem 89
Find all real zeros of the polynomial function. $$f(x)=5 x^{4}+9 x^{3}-19 x^{2}-3 x$$
View solution