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stich3 [128]
4 years ago
10

Geoff 's company manufactures customized T-shirts. Each shirt costs $2.00 to manufacture and print. The fi xed costs for this pr

oduct line are $2,000. The demand function is q = –1,200p + 7,800, where p is the price for each shirt. a. Write the expense equation in terms of the demand q. b. Express the expense equation from part a in terms of the price p. c Write the revenue function in terms of the price. d. Graph the functions in an appropriate viewing window. What price yields the maximum revenue? What is the revenue at that price? Identify the price at the breakeven points. and Round answers to the nearest cent.

Business
1 answer:
Marat540 [252]4 years ago
7 0

Answer:

  • (a) E(q) = 2,000 + 2q
  • (b) E(p) = 17,600 – 2400p
  • (c) R(p) =  –1,200p² + 7,800p

  • (d) See the graph attached:

  • The price that yields the maximum revenue is $3.25
  • The revenue at that price is $12,675
  • The prices at the breakeven points are $2.41 and $6.09

Explanation:

1. Data:

  • i) Unit variable cost: $2.00/u
  • ii) Fixed costs: $2,000.
  • iii) Demand function: q = –1,200p + 7,800  (p is the price, q is the quantity demanded)

2. Solution

<em><u>a. Write the expense equation in terms of the demand q. </u></em>

  • Expense = total cost
  • Total cost = fixed costs + variable cost
  • Variable costs = quanity × unit variable cost

  • Total cost = $2,000 + q × $2/u

  • Expense equation: E(q) = 2,000 + 2q ← answer

<em><u>b. Express the expense equation from part a in terms of the price p</u></em>

This is a composition of functions.

  • Expense equation, E(p) = E[q(p)]

  • E[q(p)] = 2,000 + 2(–1,200p + 7,800)

  • E[q(p)] = 2,000 – 2400p + 15,600 = 17,600 – 2400p

  • E[q(p)] =  17,600 – 2400p = E(p)

  • E(p) = 17,600 – 2400p

<em><u>c. Write the revenue function in terms of the price. </u></em>

Revenue is the product of the quantity demanded and the price

  • R(p) = p × q = p × [ –1,200p + 7,800]

  • R(p) =  –1,200p² + 7,800p

<u><em>d. Graph the functions in an appropriate viewing window. </em></u>

To draw the revenue function take into account:

  • It is a parabola
  • It opens downward
  • The y-intercept is (0,0) [you caluate it when p = 0]
  • The xintercpets are (0,0) and (0,6.5)
  • The vertex is at p equal to the midpoint between the two x-intercepts:

             p = (0 + 6.5)/2 = 3.25

  • What price yields the maximum revenue?

          p = 3.25

  • What is the revenue at that price?

         R(3.25) = - 1200(3.25)² + 7800(3.25) = 12,675

To draw the expense function:

  • It is a line
  • the y-intercept is (0, 17600) [make p = 0 in the equation]
  • the x-intercept is (7.33,0) [make E(p) = 0 in the equation)

  • Identify the price at the breakeven points. and Round answers to the nearest cent.

The breakeven points are when the two functions intersect. They corresponding p values are.

Thus they are the solutions to the equation:

  • -1200p² + 7800p = 17600 - 2400p

  • 1200p² -10200p + 17600 =0

Use the quadratic equation:

       x=\dfrac{10200\pm \sqrt{(-10200)^2-4(1200(17600)}}{2(1200)}

  • x = p = 2.41 and 6.09

The solutions are also shown on the graph:

  • p ≈ $2.41 and p  ≈ $6.09

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