Problem 31
Question
Solve. The cost \(C\) in dollars of manufacturing \(x\) bicycles at Holladay's Production Plant is given by the function \(C(x)=2 x^{2}-800 x+92,000\) a. Find the number of bicycles that must be manufactured to minimize the cost. b. Find the minimum cost.
Step-by-Step Solution
Verified Answer
a. 200 bicycles, b. $12,000
1Step 1: Identify the type of function
The function given is a quadratic function of the form \(C(x) = ax^2 + bx + c\), where \(a = 2\), \(b = -800\), and \(c = 92,000\). Quadratic functions have parabolic graphs which have either a minimum or maximum point, determined by the coefficient \(a\). Since \(a = 2 > 0\), the parabola opens upwards, indicating there is a minimum point.
2Step 2: Find the x-value of the vertex
The x-value of the vertex of a parabola given by the function \(ax^2 + bx + c\) can be found using the formula \(x = -\frac{b}{2a}\). Substitute \(a = 2\) and \(b = -800\) into the formula: \[ x = -\frac{-800}{2 \times 2} = \frac{800}{4} = 200. \]The number of bicycles that must be manufactured to minimize the cost is 200.
3Step 3: Calculate the minimum cost
Substitute \(x = 200\) back into the cost function \(C(x) = 2x^2 - 800x + 92,000\) to find the minimum cost:\[C(200) = 2(200)^2 - 800(200) + 92,000.\]Calculate each term:\(2(200)^2 = 80,000\)\(-800(200) = -160,000\)Now calculate the full expression:\[ C(200) = 80,000 - 160,000 + 92,000 = 12,000. \] The minimum cost is $12,000.
Key Concepts
ParabolasVertex CalculationMinimum Cost in Manufacturing
Parabolas
Parabolas are a type of curve on a graph that are part of the family of quadratic functions. These curves resemble the shape of a U or an upside-down U. The overall shape of a parabola is determined by its leading coefficient, which is the number in front of the squared term in the quadratic equation.
For example, consider the quadratic function:
For example, consider the quadratic function:
- When the leading coefficient (\( a \) ) is positive (greater than zero), the parabola opens upwards, creating the shape of a U.
- Conversely, when (\( a \) ) is negative (less than zero), the parabola opens downwards, resembling an upside-down U.
Vertex Calculation
Calculating the vertex is a key step in utilizing quadratic functions to solve practical problems. In the context of cost functions, finding the vertex helps us determine the minimum cost or resource use at a manufacturing plant, like Holladay's Production Plant.
To find the vertex of a quadratic function given by:\( ax^2 + bx + c \), we use the formula for the x-value of the vertex: \[ x = -\frac{b}{2a} \].
For instance, in the function included in the exercise: \( C(x) = 2x^2 - 800x + 92,000 \):
To find the vertex of a quadratic function given by:\( ax^2 + bx + c \), we use the formula for the x-value of the vertex: \[ x = -\frac{b}{2a} \].
For instance, in the function included in the exercise: \( C(x) = 2x^2 - 800x + 92,000 \):
- Substitute (\( a = 2 \) and \( b = -800 \)) into the vertex formula: \[ x = -\frac{-800}{2 \times 2} = \frac{800}{4} = 200 \].
- This means that the minimum cost occurs when 200 bicycles are manufactured.
Minimum Cost in Manufacturing
In many manufacturing settings, understanding how to determine the minimum cost is crucial for efficiency and savings. By using the vertex calculation derived from quadratic functions, businesses can optimize the number of goods produced to minimize costs. In the given exercise, we already determined that the minimum cost corresponds to the production of 200 bicycles.
To find the actual minimum cost, we need to substitute this x-value into the original cost function: \[ C(x) = 2x^2 - 800x + 92,000 \].
Using 200 in place of x results in:
To find the actual minimum cost, we need to substitute this x-value into the original cost function: \[ C(x) = 2x^2 - 800x + 92,000 \].
Using 200 in place of x results in:
- First, compute \( 2(200)^2 = 80,000 \).
- Next, calculate \(-800(200) = -160,000\).
- Finally, evaluate the function: \[ C(200) = 80,000 - 160,000 + 92,000 = 12,000 \].
Other exercises in this chapter
Problem 31
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