Until recently, hamburgers at the city sports arena cost $ 2.50 each. The food concessionaire sold an average of 1750 hamburgers on game night. When the price was raised to $ 3.10 hamburger sales dropped off to an average of 1,450 per night. The? concessionaire's fixed costs were $1,903.00 per night and the variable cost was $1.44 per hamburger. Answer the following questions (A) through (F).(A) Assume that the relationship between price p and demand x is linear. Express p as a function of x and find the domain of this function. p=The domain of p is(Type a compound? inequality.)
(B) Find the revenue function in terms of x and state its domain.- ?R(x)=The domain of R(x) is(Type a compound? inequality.)(C) Assume that the cost function is linear. Express the cost function in terms of x. C(x)=(D) Graph the cost function and the revenue function in the same coordinate system. Find the break-even points. The break-even points are(Simplify your answer. Type an ordered pair. Use a comma to separate answers as? needed.)(E) Find the profit function in terms of x. P(x)=(F) Evaluate the marginal profit at x=600 and interpret the results. The marginal profit at x=600 is $.b)At a production level of 600 ?hamburgers, the profit is increasing at a rate of $ per hamburger. a)At a production level of 600 ?hamburgers, the profit is decreasing at a rate of $ per hamburger.

  (A) p = -0.002x +6; 0 ≤ x ≤ 3000

  (B) R(x) = x(6 -0.002x); 0 ≤ x ≤ 3000

  (C) C(x) = 1.44x +1903

  (D) (550, 2695), (1730, 4394.20)

  (E) P(x) = -0.002x^2 +4.56x -1903

  (F) increasing at $2.16 per hamburger

Explanation:

(A) The two-point form of the equation for a line can be used.

  y = (y2 -y1)/(x2 -x1)(x -x1) +y1

The two points we have are ...

  (x, p) = {(1750, 2.50), (1450, 3.10)}

so the equation is ...

  p = (3.10 -2.50)/(1450 -1750)/(x -1750) +2.50

  p = 0.6/-300(x -1750) +2.50

  p = -0.002x +6

The domain of this function is where x and p are greater than 0. That will be for ...

  0 ≤ x ≤ 3000

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(B) Revenue is the product of burgers sold (x) and their price (p).

  R(x) = xp

  R(x) = x(6 -0.002x)

The domain of R(x) is 0 ≤ x ≤ 3000. This is the same as the domain of p(x).

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(C) The cost function is the sum of fixed costs and variable costs:

  C(x) = 1.44x +1903

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(D) See the attachment for a graph of cost and revenue. The break-even points are (x, revenue) = (550, 2695), (1730, 4394.20).

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(E) Profit is the difference between revenue and cost.

  P(x) = R(x) - C(x) = x(6 -0.002x) -(1.44x +1903)

  P(x) = -0.002x^2 +4.56x -1903

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(F) The marginal profit is the derivative of the profit function:

  P'(x) = -0.004x +4.56

  P'(600) = -0.004(600) +4.56 = -2.40 +4.56 = 2.16

At a production level of 600, the profit is increasing at a rate of $2.16 per hamburger.

whole quiz

1. the sum of all activities involved in getting high-quality products into the marketplace

2.operations

3. the principle of looking at the supply chain as a whole in order to improve the overall flow through the system.

4. produced; performed

5. activities producing tangible products, such as radios, newspapers, buses, and textbooks.

6. activities producing intangible and tangible products, such as entertainment, transportation, and education.

7.   customer focus

8. analyzing a competitor's products to identify desirable improvements

9. the systematic direction and control of the processes that transform resources into finished projects that create value for and provide benefits to consumers.

10. operation management.

your welcome leave a rate and a < 3

2b2t

explanation:

2b2t