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.

FALSE IS RIGHT

Explanation: