Problem 54 - College Algebra and Calculus: An Applied Approach

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.) \(C=312.5-10 x+0.4 x^{2} \quad C=\$ 900\)

Step-by-Step Solution

Answer

The manufacturer can produce 213 units for the cost of \$900.

Step 1 - Substitute the cost into the equation

Replace \(C\) in the equation \(C = 312.5 - 10x + 0.4x^2\) with \$900. The equation becomes \(900 = 312.5 - 10x + 0.4x^2\)

Step 2 - Rearrange the equation

Collect all terms on one side of the equation to set it equal to 0. This gives the quadratic equation \(0.4x^2 - 10x + (312.5 - 900) = 0\)

Step 3 - Simplify the equation

Simplify the equation to get \(0.4x^2 - 10x - 587.5 = 0\)

Step 4 - Solve the quadratic equation

Use the quadratic formula \(x = [-b ± sqrt(b^2 - 4ac)] / 2a\) to find the solutions. Substituting \(a = 0.4\), \(b = -10\) and \(c = -587.5\) gives two solutions:

Step 5 - Round to nearest positive integer

Round the positive solution to the nearest integer to find the number of units a manufacturer can produce for the cost \$900.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool in algebra that allows us to find the roots, or solutions, of a quadratic equation. A quadratic equation is any equation that can be written in the form \( ax^2 + bx + c = 0 \), where \(a \), \(b \), and \(c \) are constants and \(a \eq 0\).

The formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(x \) represents the variable, and \(b^2 - 4ac \) is called the discriminant. If the discriminant is positive, there are two distinct real roots. If it is zero, there is exactly one real root, also known as a repeated root. Lastly, if the discriminant is negative, it indicates that there are no real roots, only complex ones.

To apply the quadratic formula to our original exercise, we first identify the coefficients from the given cost equation, with \(a = 0.4\), \(b = -10\), and \(c = -587.5\). Plugging these into the quadratic formula gives us the potential number of units \(x \) that can be produced for a certain cost.

Cost Equation

A cost equation represents the total cost \(C\) as a function of the number of units produced \(x\). It typically includes fixed costs, variable costs per unit, and sometimes additional terms that account for factors like efficiency or scale.

In more complex scenarios, this relationship can be represented by a quadratic equation, acknowledging that costs do not always increase linearly with production. The given exercise presents such a scenario. By substituting \(C = $900 \) into the quadratic cost equation \(C = 312.5 - 10x + 0.4x^2\), we set up the equation to solve for the number of units \(x\) that lead to the specified cost.

Importance of Accurate Cost Prediction

Understanding the cost equation is crucial for businesses as it helps in making informed decisions regarding pricing, budgeting, and production planning. By calculating the break-even point or cost optimization, a company can strategize to improve its profitability.

Algebraic Problem-Solving

Algebraic problem-solving involves finding unknown values by setting up and manipulating equations. Key steps in this process include defining variables, understanding the relationships between them, and applying the appropriate algebraic techniques to solve the equations at hand.

In the context of the given exercise, the problem-solving process included substituting values into the cost equation, rearranging terms to standard quadratic form, simplifying the expression, and finally applying the quadratic formula to find the solution for the number of units \(x\).

Logical Approach to Problem-Solving

Each step taken—from the rearrangement to the application of the quadratic formula—requires a logical and methodical approach. This ensures the accuracy of the solution and increases understanding of the algebraic concepts involved. Effective problem-solving in algebra not only leads to finding solutions but also enhances critical thinking skills, which are invaluable in mathematics and beyond.