Question

Use the cost equation to find the number of units \(x\) that a manufacturer can produce for the cost \(C\). (Round your answer to the nearest positive integer.) \(C=0.125 x^{2}+20 x+5000 \quad C=\$ 14,000\)

Short answer

The manufacturer can produce approximately 600 units for the cost of $14,000.

Step-by-step solution

Setting up the equation

We are given that the cost \(C = $14000\). So we can replace \(C\) in the cost equation with 14000. That gives us: \(0.125x^2 + 20x + 5000 = 14000\)

Simplifying the equation

Subtract 14000 from both sides of the equation to set the equation to zero which will help us solve for \(x\). This gives: \(0.125x^2 + 20x + 5000 - 14000 = 0\). Simplify to get: \(0.125x^2 + 20x - 9000 = 0\)

Solving the quadratic equation

This is a quadratic equation of the form \(ax^2 + bx + c = 0\) which can be solved using the formula \(x = (-b ± sqrt(b^2 - 4ac)) / (2a)\). Plugging the values \(a = 0.125\), \(b = 20\), and \(c = -9000\) into the formula, we get: \(x = [ -20 ± sqrt( (20)^2 - 4*0.125*(-9000) ) ] / (2*0.125) \)

Calculating the value of \(x\)

After calculating the expression inside the square root and simplifying the equation we get two solutions. But as \(x\) represents the number of units produced, it must be a positive number, thus we reject the negative solution. Rounding our answer to the nearest integer, we get: \(x = 600\)

Key concepts

Cost Equation

Writing a cost equation is an essential part of understanding how production expenses relate to the number of units produced. In the given exercise, you start with the cost equation: \[ \text{C}(x) = 0.125x^2 + 20x + 5000 \]This equation includes three components:

• Quadratic Term (\(0.125x^2\)): Represents costs that increase with the square of the production quantity. These often include scalable expenses such as power consumption.
• Linear Term (\(20x\)): This encompasses variable costs tied to production, typically labor and materials.
• Constant Term (5000): These are fixed costs that remain constant regardless of the number of units produced, like rent and utilities.

By substituting different values into the equation, you can calculate the total cost for producing any specific number of units. This powerful tool helps in decision-making and forecasting costs accurately.

Solving Quadratic Equations

Quadratic equations are essential in many fields to find unknown values. Here, solving the quadratic equation is crucial for determining the production capacity given a specific cost.To solve a quadratic equation of the form:\[ ax^2 + bx + c = 0 \]We typically use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In our context:

• We set up the equation by replacing the cost \(C\) with \(14,000\), giving us a working equation of \(0.125x^2 + 20x + 5000 = 14000\).
• Simplifying the equation to make it a standard quadratic form, we translate it to \(0.125x^2 + 20x - 9000 = 0\).
• Now, apply the quadratic formula using \(a = 0.125\), \(b = 20\), and \(c = -9000\).

Solving proceeds with calculating the discriminant \(b^2 - 4ac\) and finding the values of \(x\). Always verify the feasibility of solutions within the context, such as ignoring negative units produced.

Production Quantity Calculation

Accurately calculating production quantity is vital for manufacturers aiming to stay within budget while meeting demand.In this example, once the quadratic equation is simplified and applied in the cost context:

• Use the derived formula to determine potential production scenarios that align with given cost constraints.
• Solve to find possible solutions of \(x\), which represent units produced.
• Recognize realistic values, as negative production does not make sense in most practical applications. So, you only consider the positive solution.

Ultimately, after applying the calculations and rounding to the nearest positive integer, we find that producing 600 units meets the cost condition of \(C = 14000\). This process demonstrates how mathematical concepts directly impact real-world industrial strategies.