What will the profit be if 1500 units are sold next year?

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Discussion Overview

The discussion centers around calculating the profit for a small company if 1500 units are sold in the upcoming year. It involves exploring profit functions, cost structures, and different methods of analysis related to financial forecasting.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant outlines the profit function as \( P(p,x) = (p-80)x - 15000 \), indicating fixed costs and marginal costs.
  • Another participant questions the method by suggesting an alternative approach involving a chart to organize unit costs and profits.
  • A later reply proposes a simpler algebraic method to derive the profit function, emphasizing efficiency over the chart method.
  • There is an implicit assumption that the selling price \( p \) remains constant from the previous year.

Areas of Agreement / Disagreement

Participants express differing views on the best method to calculate profit, with no consensus on which approach is superior. The discussion remains unresolved regarding the most effective calculation method.

Contextual Notes

Limitations include the lack of clarity on the selling price \( p \) and the assumptions regarding fixed and variable costs. The discussion does not resolve how these factors influence the profit calculation.

Simon T
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the first year of operation for a small company yielded a loss of \$5000. The company has \$15,000 per year tied up in fixed costs and spent 40,000 on raw materials and labour. Since the company was not well known, they were only able to sell 500 units of their product that year. It is hoped that business with improve next year and that 1500 units will be sold at the same selling price as last year.

a) If 1500 units are sold next year, what will the profit be?
 
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Profit is revenue less costs. There is a fixed cost of 15000 and a marginal cost of 80...if we denote the price of each unit as $p$ and the number of units sold as $x$, then our profit function $P$ is:

$$P(p,x)=px-\left(80x+15000\right)=(p-80)x-15000$$

Now, we are given:

$$P(p,500)=(p-80)500-15000=-5000$$

So, what is $p$?
 
I was taught a different way for this module. I was taught to have a chart with Unit basis and total amount at the top and on the left side we have s, vc, cm, fc, and np and we fill in the numbers.
 
Well, you're in for a treat then, as I'm going to help you answer this question in a much easier and more straightforward manner using a little algebra. (Yes)

Once you find $p$, then you will have the profit function in one variable $x$, into which you can simply plug the given value of $x$ to output the profit for that production level. No time consuming charts, graphs and/or tables required. (Bow)
 

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