Problem 48 - Calculus Volume 1

Question

[T] The total cost \(C(x),\) in hundreds of dollars, to produce \(x\) jars of mayonnaise is given by \(C(x)=0.000003 x^{3}+4 x+300\) a. Calculate the average cost per jar over the following intervals: i. \([100,100.1]\) ii. \([100,100.01]\) iii. \([100,100.001]\) iv. \([100,100.0001]\) b. Use the answers from a. to estimate the average cost to produce 100 jars of mayonnaise.

Step-by-Step Solution

Answer

The average cost to produce 100 jars is estimated to be $40 per jar.

Step 1: Calculate average cost formula

The average cost function over an interval \([a, b]\) is given by \(\frac{C(b) - C(a)}{b - a}\). We'll use this formula to calculate the average cost over the given intervals.

Step 2: Calculate average cost for interval [100,100.1]

For \([100, 100.1]\): - First, calculate \(C(100) = 0.000003 \times 100^3 + 4 \times 100 + 300 = 700\) hundreds of dollars. - Next, calculate \(C(100.1) = 0.000003 \times 100.1^3 + 4 \times 100.1 + 300 \approx 700.04003\) hundreds of dollars. - The average cost is \(\frac{700.04003 - 700}{100.1 - 100} = 0.4003\) hundreds of dollars per jar.

Step 3: Calculate average cost for interval [100,100.01]

For \([100, 100.01]\): - Calculate \(C(100.01) = 0.000003 \times 100.01^3 + 4 \times 100.01 + 300 \approx 700.0040003\) hundreds of dollars. - The average cost is \(\frac{700.0040003 - 700}{100.01 - 100} = 0.40003\) hundreds of dollars per jar.

Step 4: Calculate average cost for interval [100,100.001]

For \([100, 100.001]\): - Calculate \(C(100.001) = 0.000003 \times 100.001^3 + 4 \times 100.001 + 300 \approx 700.000400003\) hundreds of dollars. - The average cost is \(\frac{700.000400003 - 700}{100.001 - 100} = 0.400003\) hundreds of dollars per jar.

Step 5: Calculate average cost for interval [100,100.0001]

For \([100, 100.0001]\): - Calculate \(C(100.0001) = 0.000003 \times 100.0001^3 + 4 \times 100.0001 + 300 \approx 700.0000400003\) hundreds of dollars. - The average cost is \(\frac{700.0000400003 - 700}{100.0001 - 100} = 0.4000003\) hundreds of dollars per jar.

Step 6: Estimate the average cost for 100 jars

Based on the calculated values for smaller and smaller intervals, the estimates show that the average cost per jar of producing 100 jars is approaching \(0.4\) hundreds of dollars, or \(40\) dollars per jar.

Key Concepts

Interval Analysis

Interval analysis helps us understand how a function behaves over a specific range of values. In the context of cost functions, this means examining how the cost per unit changes as we slightly alter the quantity. By breaking down the problem into smaller intervals, you can find the average cost per jar over these tiny ranges.

Consider different intervals like \(100, 100.1\) or \(100, 100.001\). Calculating the cost over these shrinking intervals helps us grasp the behavior of the cost function at a specific point. This process provides more reliable estimates for financial decisions. As intervals get tinier, they offer a closer look at what happens at a specific value, aiding in making accurate predictions.

Cost Function

The cost function, denoted by \(C(x)\), represents the total cost of producing \(x\) units of a product. It's essentially a mathematical expression that breaks down costs into fixed and variable components. In this scenario, \(C(x) = 0.000003x^3 + 4x + 300\) takes a cube and linear term into account, along with a fixed cost of 300.

The cube term \(0.000003x^3\) suggests that cost increases at a growing rate as production scales up. Meanwhile, the linear term \(4x\) and the constant term \(300\) cover typical production costs. Understanding how each part affects the overall expense can help you better manage manufacturing efficiency and pricing strategies.

Limit of a Function

The limit helps us find the value that a function approaches as the input approaches a specific point. In calculating average costs, we use limits to determine the precise cost as intervals become infinitesimally small.

The process of reducing the size of intervals helps identify the limit of the average cost function. As intervals shrink around a particular value, the average cost becomes more accurate and predictable. This method allows us to pinpoint precise costs at any given production level, ensuring strategic financial decision-making.

Derivatives

Derivatives indicate how a function changes at any point, reflecting the rate at which this change occurs. In cost analysis, they reveal important insights into how costs fluctuate with production volumes. Calculating a derivative involves finding how a small change in quantity \(x\) affects the cost \(C(x)\).

The derivative of the cost function here would represent the marginal cost of increasing production by one jar. It can be obtained through calculus, ensuring that businesses remain efficient and cost-effective. By understanding this, companies can streamline operations and optimize resource allocation, responding swiftly to market demands.