Answer:
(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.
