A tech company has developed a new compact, high efficiency battery for hand-held devices.

Sample Solution

Part 1: Breakeven Point with Substitution Method

Given the graphs representing cost and revenue of battery production, we need to find the point where both values are equal, essentially the breakeven point. We can achieve this using the substitution method.

1. Identify Equations from Graph: From the graph, we can identify the cost and revenue equations:

• Cost: C(x) = 3x + 0.5 (linear equation)
• Revenue: R(x) = -2x^2 + 10x + 7 (quadratic equation)

2. Set Equations Equal: To find the point where cost and revenue are equal, we set the cost function equal to the revenue function:

Full Answer Section

C(x) = R(x) 3x + 0.5 = -2x^2 + 10x + 7

3. Solve for x: This equation becomes a quadratic equation. We can solve it by:

• Moving the constant term to the right side: 2x^2 - 7x - 6.5 = 0
• Factoring the equation: (x - 2)(2x + 3.25) = 0
• Finding the possible solutions: x = 2, x = -\frac{3.25}{2}

Part 2: Interpretation of Results

From the solutions, we have two potential breakeven points:

• x = 2: This indicates producing 2 thousand batteries reaches the breakeven point, where cost and revenue are equal.
• x = -1.625: This solution is negative and doesn't make sense in the context of producing batteries. Therefore, the valid breakeven point occurs at 2 thousand batteries.

Part 3: Correspondence with the Graph

Yes, our result coincides with the graph. Looking at the intersection point of the cost and revenue curves, we find it indeed occurs at x = 2, aligning with the production of 2 thousand batteries.

Part 4: Profit Equation

Profit is the difference between revenue and cost. Therefore, the profit equation can be written as:

P(x) = R(x) - C(x)

Substituting the revenue and cost equations:

P(x) = (-2x^2 + 10x + 7) - (3x + 0.5) P(x) = -2x^2 + 7x + 6.5

This equation represents the profit earned at any production level (x).

Part 5: Profit at 100 Thousand Batteries

To find the profit from producing 100 thousand batteries (x = 100), we plug this value into the profit equation:

P(100) = -2(100)^2 + 7(100) + 6.5 P(100) = -20000 + 700 + 6.5 P(100) = -12293.5

However, the negative profit value indicates a loss of 12,293.5 units at the production level of 100 thousand batteries.

Conclusion:

By utilizing the substitution method and understanding the cost and revenue functions, we identified the breakeven point at 2 thousand batteries and derived the profit equation. However, at 100 thousand batteries, the predicted outcome is a loss, highlighting the importance of considering production volume and profit margins for sustainable operations.