Problem 52 - 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.) \(C=0.5 x^{2}+15 x+5000 \quad C=\$ 11,500\)

Step-by-Step Solution

Answer

The manufacturer can produce approximately 354 units for a cost of $11,500.

Step 1: Understand the problem

We need to find the value of \(x\) when the cost \(C\) is \$11,500. That means we set \(C\) equal to 11,500 in the cost equation \(C=0.5 x^{2}+15 x+5000\). The equation will be \(0.5x^2+15x+5000=11500\).

Step 2: Rearrange the equation

To isolate \(x\), first we move \(11500\) to the other side so we get \(0.5x^2+15x+5000-11500=0\), which simplifies to \(0.5x^2 + 15x - 6500 = 0\). Then we divide by \(0.5\) to remove the coefficient from the \(x^2\) term, the simplified equation is \(x^2 + 30x -13000 = 0\).

Step 3: Compute the roots using the quadratic formula

The quadratic formula for a quadratic equation ax\(^2\) + bx + c = 0 is\( x= {-b \pm \sqrt{b^2-4ac}}/{2a} \). Substituting a=1, b=30 and c=-13000 into the formula, we get two roots.

Step 4: Interpret the solution

There are two roots from step 3, but in the context of this problem, \(x\) represents the number of units, which must be a positive integer. Thus, the answer will be the positive root of our computed values, rounded to the nearest integer.

Key Concepts

Quadratic Formula

Understanding the quadratic formula is crucial to solving quadratic equations. It is the solution to the general form of a quadratic equation, which is given by \( ax^2 + bx + c = 0 \). The formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) provides the values of x that satisfy the equation.

To apply the quadratic formula, identify the coefficients \(a\), \(b\), and \(c\) from the equation. In our example, after simplifying the cost equation for \(C = 11500\), we identified \(a = 1\), \(b = 30\), and \(c = -13000\). Plugging these into the formula gives us the roots of the quadratic equation. Remember, the \pm symbol indicates there will be two roots. One from using a plus sign, and another from using a minus sign in the equation. It's the value inside the square root, \(b^2-4ac\), called the discriminant, that can tell us about the nature of the roots before we even compute them.

Solving Quadratic Equations

When we're faced with a quadratic equation, our goal is to find the values of x that make the equation true. There are several methods to solve such equations, and one of the most common methods is by using the quadratic formula as mentioned above.

However, depending on the specific equation, other methods might be more appropriate, such as factoring, completing the square, or graphing. Each method has its own advantages. Factoring is quick and efficient if the solution is apparent, while completing the square can offer a more systematic approach. Graphing provides a visual representation, which can be useful to understand the roots' relationship with the x-axis.

Selecting the Correct Root

After solving for x using the quadratic formula, we may end up with two solutions. It's important to consider the context of the problem to determine which root is the correct one. As in our example, only the positive integer value of x as the number of units manufactured is sensible, as negative or fractional units would not be realistic.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations without an equality sign. They are the building blocks of algebraic equations. For example, in the cost equation \( C = 0.5 x^2 + 15x + 5000 \), the right side is an algebraic expression representing the cost \( C \) depending on the number of units produced \( x \).

The ability to manipulate algebraic expressions is essential for solving equations. This includes simplifying expressions by combining like terms, factoring to reveal roots or zeros, and expanding expressions using distributive properties.

Expression Simplification

Simplifying the expression often makes the problem easier to solve. In our case, dividing the quadratic equation by 0.5 simplified the coefficients, which made it more straightforward to apply the quadratic formula. Through these manipulations, we can transform complex problems into simpler ones that still conform to the same rules of algebra.