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Kani is running a business manufacturing car batteries. She notices that her profits are lower than last year and wonders why. Would her profits increase if she just sold a few more batteries, or does she need to increase the price? If she keeps prices the same, will her profits continue to drop, and will she be able to afford to keep operating? She decides to model the costs and revenues of the business so she can answer some of these questions. Kani knows that she can model the cost of running her business with the linear function C(x)=30x+600 , where x represents the number of batteries she sells. This linear function shows that she spends $30 for each battery she manufactures. She has a fixed cost of $600 regardless of the number of batteries she makes. Her financial officer tells her that the revenue is the amount of money generated by the business. Revenue can be modeled by the polynomial function R(x)=−3x^2+33x , where x represents the number of batteries sold. The total profit of the company is the revenue minus the costs. Kani needs to subtract C(x) from R(x) . She decides that the profits can be modeled by the function P(x)=−3x^2+3x−600 . Knowing this, Kani can figure out what her profit will be for different values of x . She estimates profits based on the previous year's sales. If she graphs this function, she might notice different patterns that can inform her business choices going forward. This polynomial function models a complex, real-world situation. By the end of this unit, you will be able to evaluate polynomial functions at specific values, graph polynomial functions, notice patterns, and interpret the meaning of various points on these graphs.

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Kani has a solid foundation for analyzing her business using the functions for cost, revenue, and profit. Let’s break down the problem and understand how to evaluate profits and analyze the implications of different sales scenarios.

Given Functions:

  1. Cost Function: \( C(x) = 30x + 600 \)

    • This function indicates that for every battery produced, Kani spends $30, and she has a fixed cost of $600.

  2. Revenue Function: \( R(x) = -3x^2 + 33x \)

    • This function shows the revenue generated from selling \( x \) batteries. The negative quadratic term suggests that there is a maximum revenue point, indicating diminishing returns as sales increase.

  3. Profit Function: \( P(x) = R(x) - C(x) = -3x^2 + 33x - (30x + 600) \)

    • Simplifying gives: \( P(x) = -3x^2 + 3x - 600 \)

Analyzing the Profit Function:

To understand Kani's profits, we can evaluate the profit function at specific values of \( x \) (number of batteries sold).

  1. Zero Profit Points: To find out when Kani breaks even (i.e., when profit is zero), we set the profit equation equal to zero: \[ -3x^2 + 3x - 600 = 0 \] This can be simplified by dividing through by -3: \[ x^2 - x + 200 = 0 \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(200)}}{2(1)} \ x = \frac{1 \pm \sqrt{1 - 800}}{2} \] Since \( 1 - 800 \) is negative, this means there are no real solutions to the equation; thus, \( P(x) \) does not reach zero, which indicates that Kani is currently operating at a loss.

  2. Maximum Profit: The profit function is a downward-opening parabola due to the negative coefficient in front of the \( x^2 \) term. The vertex of the parabola gives the maximum profit. The vertex formula is given by \( x = -\frac{b}{2a} \), where \( a = -3 \) and \( b = 3 \): \[ x = -\frac{3}{2(-3)} = \frac{1}{2} \] Kani should ideally sell about 0.5 batteries to maximize profit, which does not make practical sense in the context of selling whole batteries.

  3. Analyzing Specific Values of \( x \): Kani can evaluate the profit function for various values of \( x \) (like 10, 20, 30 batteries, etc.) to see how her profits change:

    • For instance, if she sells 10 batteries: \[ P(10) = -3(10^2) + 3(10) - 600 \ = -300 + 30 - 600 = -870 \ (loss) \]
    • If she sells 20 batteries: \[ P(20) = -3(20^2) + 3(20) - 600 \ = -1200 + 60 - 600 = -1740 \ (loss) \]

Conclusion:

Kani's profits are negative for the values of batteries sold that are reasonable for a business, indicating that she is operating at a loss. To improve her situation, Kani can consider:

  1. Increasing Sales: Selling more batteries can sometimes cover fixed costs and help reduce losses, especially if cost per unit decreases with volume.

  2. Increasing Prices: Raising the sale price of the batteries could directly improve revenue, but she must consider the price elasticity of demand (how price changes impact sales volume).

  3. Cost Management: Finding ways to reduce fixed or variable costs could also help increase profit margins.

By analyzing different values and making adjustments based on her findings and market conditions, Kani can strategize effectively to turn her business around.