Problem 63 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

It is estimated that the monthly cost of producing \(x\) units of a particular commodity is \(C(x)=0.06 x+3 x^{1 / 2}+20\) hundred dollars. Suppose production is decreasing at the rate of 11 units per month when the monthly production is 2,500 units. At what rate is the cost changing at this level of production?

Step-by-Step Solution

Answer

The cost is decreasing at a rate of 0.99 hundred dollars per month.

Step 1: Identify Given Information

The cost function is given as \( C(x) = 0.06x + 3x^{1/2} + 20 \) hundred dollars. The monthly production rate is decreasing at 11 units per month, and current production level is 2,500 units.

Step 2: Differentiate the Cost Function

To find how fast the cost is changing, compute the derivative of the cost function with respect to \(x\): \[ C'(x) = \frac{d}{dx}(0.06x + 3x^{1/2} + 20) \].

Step 3: Apply the Power Rule

Derive each term individually: \[ C'(x) = 0.06 + 3\left(\frac{1}{2}x^{-1/2}\right) \]. Simplify to get: \[ C'(x) = 0.06 + \frac{3}{2}x^{-1/2} \] or \[ C'(x) = 0.06 + \frac{3}{2\sqrt{x}} \].

Step 4: Substitute the Current Production Level

Substitute \(x = 2500\) into the derivative to find \(C'(2500)\): \[ C'(2500) = 0.06 + \frac{3}{2\sqrt{2500}} \].

Step 5: Simplify the Expression

Calculate the value: \[ \sqrt{2500} = 50 \], so \[ C'(2500) = 0.06 + \frac{3}{2 \times 50} = 0.06 + \frac{3}{100} = 0.06 + 0.03 = 0.09 \]. Therefore, the rate of change of the cost with respect to production is 0.09 hundred dollars per unit.

Step 6: Multiply by the Rate of Decreasing Production

Since the production is decreasing at 11 units per month, multiply: \[ \frac{dC}{dt} = C'(x) \cdot \frac{dx}{dt} = 0.09 \times (-11) \].

Step 7: Calculate Final Rate

Compute the product: \[ \frac{dC}{dt} = 0.09 \times (-11) = -0.99 \] hundred dollars per month.

Key Concepts

Cost Function

The cost function, denoted as \(C(x)\), represents the total cost of producing \(x\) units of a given commodity. In this exercise, the cost function is given as \( C(x) = 0.06x + 3x^{1/2} + 20 \). This means the cost to produce \(x\) units is determined by three components: \(0.06x\) (a linear term), \(3x^{1/2}\) (a square root term), and \(20\) (a constant term). The cost is expressed in hundreds of dollars. The function aids in understanding how production levels impact overall costs.

Derivative

A derivative, in calculus, measures how a function changes as its input changes. It's the rate at which one quantity changes with respect to another. For the cost function \(C(x)\), the derivative, noted as \(C'(x)\), gives us the rate of change of cost with respect to the number of units produced. To find \(C'(x)\), you need to differentiate \(C(x)\) with respect to \(x\). This step involves computing the derivative of each term in the cost function.

Rate of Change

The rate of change tells us how one quantity changes as another quantity changes. For this exercise, we need to find how fast the cost changes as the production level changes. By taking the derivative \(C'(x)\), we understand the rate at which cost changes for each unit of production. Subsequently, multiplying \(C'(x)\) by the rate at which production is decreasing (given as 11 units per month) tells us the overall rate at which cost is changing over time.

Power Rule

The power rule is a basic principle in calculus used to differentiate functions of the form \(x^n\). For any function \(x^n\), the derivative is given by \(nx^{n-1}\). In our cost function exercise, the power rule helps compute the derivative of \(3x^{1/2}\). Applying the power rule, the derivative of \(3x^{1/2}\) is \(3 \cdot (1/2)x^{(-1/2)} = \frac{3}{2}x^{-1/2}\). This simplification is crucial for finding the overall derivative of the cost function.

Application of Derivatives

To fully apply the derivative, we substitute the given production level (\(x = 2500\) units) into the derived expression. This gives the specific rate of change in cost at that production level. After simplification, we find \(C'(2500) = 0.09\) hundred dollars per unit. Since production is decreasing at 11 units per month, we multiply the rate of change by -11 to find the rate at which cost is changing: \( \frac{dC}{dt} = 0.09 \times -11 = -0.99 \) hundred dollars per month. This negative value indicates that the cost is decreasing as production decreases.