Problem 82 - Algebra and Trigonometry

Question

A small-appliance manufacturer finds that the profit \(P\) (in dollars) generated by producing \(x\) microwave ovens per week is given by the formula \(P=\frac{1}{10} x(300-x)\) provided that \(0 \leq x \leq 200\) . How many ovens must be manufactured in a given week to generate a profit of \(\$ 1250 ?\)

Step-by-Step Solution

Answer

Manufacture 50 ovens to generate a profit of \$1250.

Step 1: Understand the Problem

We need to determine the value of \(x\) (number of ovens) that satisfies the equation \(P = \frac{1}{10} x (300 - x)\) when the profit \(P\) is \$1250.

Step 2: Set Up the Equation

Given \(P = \$1250\), we set this equal to the profit equation: \(1250 = \frac{1}{10} x (300 - x)\).

Step 3: Clear the Fraction

Multiply both sides of the equation by 10 to eliminate the fraction: \(12500 = x (300 - x)\).

Step 4: Expand and Rearrange the Equation

Expand the right side and rearrange all terms: \(12500 = 300x - x^2\) becomes \(x^2 - 300x + 12500 = 0\).

Step 5: Solve the Quadratic Equation

Solve the quadratic equation \(x^2 - 300x + 12500 = 0\) using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -300\), \(c = 12500\).

Step 6: Calculate Discriminant

Compute the discriminant: \(b^2 - 4ac = (-300)^2 - 4(1)(12500) = 90000 - 50000 = 40000\).

Step 7: Apply the Quadratic Formula

Calculate \(x\) values: \(x = \frac{-(-300) \pm \sqrt{40000}}{2(1)} = \frac{300 \pm 200}{2}\).

Step 8: Compute the Roots

Calculate the two possible values of \(x\): \(x = \frac{300 + 200}{2} = 250\) and \(x = \frac{300 - 200}{2} = 50\).

Step 9: Verify Within Constraints

Ensure the solutions are within the constraints \(0 \leq x \leq 200\). The value \(x = 250\) is out of bounds, while \(x = 50\) is within bounds.

Key Concepts

Profit Maximization

Profit maximization is a critical goal in any production process. It involves determining the number of units to produce to achieve the highest possible profit. In our exercise, the profit from producing microwave ovens is described by a quadratic equation:

• The formula given is: \(P = \frac{1}{10} x (300 - x)\).
• Maximizing profit means finding the optimal number of ovens \(x\) that gives the desired profit of $1250.

This type of problem often involves setting the profit equal to a known value and solving for the production level that yields that profit.
Understanding the shape of the quadratic equation can also be helpful. Quadratics express relationships that form a parabola, and in this case, the parabola opens downward, indicating there's a peak point which represents maximum profit.

Microwave Ovens Production

Microwave ovens production in this context requires manipulation of the given quadratic profit formula based on production constraints. The manufacturer must decide the number of ovens to produce, represented by \(x\):

• Production should meet demand, which affects profitability.
• Constraints must be adhered to, such as: \(0 \leq x \leq 200\).

It’s crucial to balance production numbers to avoid overproduction, which leads to increased costs and inefficiencies. Insight into production levels is vital, as it directly influences profit by aligning with market needs and operational capacity.
Manufacturers should use sensitivity analysis to adapt production levels to changing profit scenarios and market demands, ensuring continuous profit optimization.

Discriminant Calculation

Discriminant calculation is a key step in solving quadratic equations, helping determine the nature of the solutions. The discriminant is found in the quadratic formula:

• For our equation, it's computed as: \(D = b^2 - 4ac\).
• Calculated here as \(90000 - 50000 = 40000\).

The discriminant tells us several things:

• If \(D > 0\), there are two real and distinct solutions.
• If \(D = 0\), there is exactly one real solution.
• If \(D < 0\), there are no real solutions.

In this exercise, a positive discriminant indicates that there are two potential production levels that might satisfy the profit condition. Therefore, verifying these values according to constraints will confirm the feasible production amount.

Constraints Verification

Constraints verification is fundamental to ensure solutions are practical and within logical bounds. After obtaining potential solutions from the quadratic equation:

• Results include \(x = 250\) and \(x = 50\).
• Constraints dictate \(0 \leq x \leq 200\), so \(x = 250\) is invalid.

Verification ensures that all other operational constraints are respected. In this problem, only \(x = 50\) is a valid solution as it fits within the allowable range. Always check computation results against real-world constraints, as theoretical solutions might not always align with practical limitations.
This approach guarantees that the selected production level is not just mathematically correct but also feasible and economically viable, aligning with business goals and operational capabilities.