Answer:
$62,800
Explanation:
Following Garfield Corp's policy, the number of pet beds that must be purchased, assuming no initial inventory, is given by the expected number of sales in March (1,300 units) added to 30% of the expected sales in April (30% of 900 units):
Since the company purchases each pet bed for $40, total budgeted purchases are:
Garfield Corp's total budgeted purchases for March are $62,800.
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.
