Math 133 Unit 2 Individual Project 2b Student Answer Form

Math133 Unit 2 Individual Project 2b Student Answer Formname Require

Suppose you have a start-up company that develops and sells a gaming app for smartphones. You get to know the financial performance of this by understanding your cost, revenue, and profit. The monthly cost function (in US dollars) of developing your app is C(x) = 3x + b, where C(x) is the cost and x is the number of app downloads. The $3 in the equation is called the variable cost per unit, and b is called the fixed cost.

The monthly revenue function (in US dollars) based on previous monthly sales is modeled by the function R(x) = -0.15x2 + 153x, where 0 ≤ x ≤ 500. The monthly profit function (in US dollars), P(x), is derived by subtracting the cost from the revenue in the function P(x) = R(x) – C(x). For each question, be sure to show all your work details for full credit. When applicable, round all answers to the nearest cent and whole download values.

Paper For Above instruction

In this project, we analyze the financial aspects of a start-up company developing a gaming app for smartphones. The focus is on understanding how costs, revenue, and profit change based on the number of downloads, with particular attention to how fixed costs influence overall profitability.

Question 1: Based on the first letter of your last name, choose a value for your fixed cost, b. The options are:

  • A–F: $1,000–$1,500
  • G–L: $1,501–$2,000
  • M–R: $2,001–$2,500
  • S–Z: $2,501–$3,000

Suppose, for example, that my last name begins with M, which falls in the range M–R. I select a fixed cost of $2,200 within this interval. Thus, the fixed cost b = $2,200.

Using this fixed cost, the cost function becomes:

C(x) = 3x + 2200

This means that the total monthly development cost depends linearly on the number of downloads, with a fixed initial investment of $2200.

Question 2: State the profit function.

Recall that profit P(x) is calculated as revenue minus cost:

P(x) = R(x) – C(x)

Given R(x) = -0.15x2 + 153x and C(x) = 3x + 2200, the profit function becomes:

P(x) = (-0.15x2 + 153x) – (3x + 2200)

which simplifies to:

P(x) = -0.15x2 + 153x – 3x – 2200 = -0.15x2 + 150x – 2200

This quadratic function models profit depending on the number of downloads, x, within the range 0 to 500.

Question 3: Show calculation details.

To derive the profit function, start with the given revenue and cost functions:

  • Revenue: R(x) = -0.15x2 + 153x
  • Cost: C(x) = 3x + 2200

Subtract C(x) from R(x):

P(x) = R(x) – C(x) = [-0.15x2 + 153x] – [3x + 2200]

Distribute the negative sign in the second bracket:

P(x) = -0.15x2 + 153x – 3x – 2200

Combine like terms:

P(x) = -0.15x2 + 150x – 2200

This is the simplified profit function depending on the number of downloads.

Question 4: Show calculation details.

To determine the profit at specific download numbers, substitute the values into the profit function. For example, at x=0 (no downloads):

P(0) = -0.15(0)2 + 150(0) – 2200 = -2200

This indicates a loss equal to the fixed costs when no downloads occur, as expected.

At x=250, the midpoint of the range:

P(250) = -0.15(250)2 + 150(250) – 2200

= -0.15(62,500) + 37,500 – 2200

= -9,375 + 37,500 – 2200 = 25,925

This suggests significant profit at 250 downloads, demonstrating how increasing downloads enhances profitability.

Question 5: Show calculation details for the profit function. x, number of downloads, P(x), profit in US dollars.

Using the profit function P(x) = -0.15x2 + 150x – 2200, we analyze key points:

  • Maximum profit: Since P(x) is quadratic with a negative leading coefficient, it has a maximum at x = –b / (2a), where a = -0.15 and b = 150.

x = –(150) / (2 * -0.15) = –150 / -0.3 = 500

At x=500, the maximum number of downloads within the range, profit is:

P(500) = -0.15(500)2 + 150(500) – 2200

= -0.15(250,000) + 75,000 – 2200 = -37,500 + 75,000 – 2200 = 35,300

Thus, the maximum profit of approximately $35,300 occurs at 500 downloads.

Graphically, the parabola opens downward, with the vertex at x=500, indicating the peak profit point.

Question 6: Show the graph. (Note: Drawing a graph isn't possible here, but you should plot P(x) from x=0 to 500 with key points such as x=0, 250, and 500, and indicate the vertex at x=500.)

In a typical graph, the profit curve starts at -2200 when no downloads occur, rises to a maximum at 500 downloads, and is symmetrical around the vertex. Plotting the key points confirms the parabola's shape and the profit's dependence on downloads.

Question 7: Apply critical thinking to answer this question.

Analyzing the profit function reveals that increasing downloads initially improves profit, up to the maximum at 500 downloads. This suggests that marketing efforts should focus on reaching the maximum target downloads. However, beyond 500 downloads, the model is only valid up to this range; in reality, constraints such as market saturation may limit downloads.

Furthermore, fixed costs significantly influence profitability. A higher fixed cost reduces overall profit, emphasizing the importance of controlling development expenses. It also indicates that to break even, the number of downloads must be sufficient to offset fixed costs and variable costs.

From a strategic perspective, understanding the optimal number of downloads for maximum profit can help in planning marketing budgets, setting realistic sales targets, and evaluating app performance relative to costs. Additionally, as the revenue function is quadratic with a downward opening, increasing downloads beyond certain limits won't continue to increase profits and may lead to losses if fixed costs aren't recovered early.

In conclusion, analyzing the profit function underscores the importance of balancing fixed and variable costs, setting realistic sales goals, and understanding market saturation effects. These factors are critical for making informed decisions in the app development business to optimize profitability.

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