ClassHW_1

# ClassHW_1 - Hw-1 Solutions 1(2-4 Cost Rent Hauling Total...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Hw-1 Solutions 1. (2-4) Cost Rent Hauling Total Cost Site A = \$5,000 (4)(200,000)(\$1.50) = \$1,200,000 \$1,205,000 Site B = \$100,000 (3)(200,000)(\$1.50) = \$900,000 \$1,000,000 Note that the revenue of \$8.00/yd3 is independent of the site selected. Thus, we can maximize profit by minimizing total cost. The solid waste site should be located in Site B. 2. (2-12) (a) p = 150 - 0.02D v = 53 f = 42,000 π = Total Revenue - Total Cost = (150 - 0.02D)D - (42,000 + 53D) = 97D - 0.02D2 - 42,000 D* = 2425 units Since the second derivative is negative, profit has been maximized at D*. (b) Max Profit = Profit(D=2425) = 97(2425) – 0.02(2425)2 – 42,000 = \$75,612.50 (c) Profit = 97D - 0.02D2 - 42,000 = 0 Solving quadratically, we obtain two solutions: D = 481, or, D = 4,369. Our maximum output is 4,000, so the only breakeven is at D = 481. (d) The range of profitable demand is (481, 4,000]. Breakeven occurs at 481 units, and maximum output is 4000. So by (c), producing in this range ensures profit. 3. (2-14) (a) p = 600 - 0.05D f = 900,000 v = 131.50 π = Total Revenue - Total Cost = (600 - 0.05D)D - (900,000 + 131.5D) = 468.5D – 0.05D2 - 900,000 dπ/dD = 468.5 – 0.1D D* = 4,685 units per month π = Profit(D=4685) = \$197,461 (b) Find the range of values for D such that π > 0. Solving 468.5D – 0.05D2 - 900,000 = 0 gives us two values, D = 2,698 and D = 6,672. D Sign 2,698 — 6,672 + — Therefore, domain of profitable demand is (2698, 6672). 4. (2-15) (a) p = 38 + 2700/D – 5000/D2 f = 1,000 v = 40 π = Total Revenue – Total Cost = (38 + 2700/D – 5000/D2)D – (1000 + 40D) = -2D + 1700 – 5000/D dπ/dD = –2 + 5000/D2 D* = 50 units per month (b) d2π/dD2 = –10000/D3 d2π/dD2(D=50) = –0.08 < 0 Therefore, at D = 50, profit is maximized. 5. (2-17) Breakeven point in units of production: f = \$100,000/yr ; Cv = \$140,000/yr (70% of capacity) Sales = \$280,000/yr (70% of capacity); p = \$40/unit Annual Sales (units) = \$280,000/\$40 = 7,000 units/yr (70% capacity) v = \$140,000/7,000 = \$20/unit Breakeven point: TR = TC pD = f+ vD 40D = 100,000 + 20D D = 5,000 units/year, or 50% utilization of capacity 6. (2-25) Let x = the number of weeks to wait. x = 0, 1, 2, … Quantity(x) = 1,000 + 1,200x – 200x Price(x) = \$3.00 - 0.50x π = (1,000 + 1,200x – 200x)(3 – 0.5x) = 3,000 + 2,500x – 500x2 x* = 2.5 weeks Profit = \$6,125 7. π = x3/3 – 5 x2 + 16x + 20 where x = amount of products produced in one day. The domain of x is [0, 12]. = x2 – 10x + 16 = 0 (x – 8)(x – 2) = 0 Local optima exist at x = 2 and x = 8. The global maximum is found by evaluating the profit function at the local optima and at the endpoints of the domain of x. x π 0 20 2 29.33 8 -1.33 The firm should manufacture 12 units of product each day. 12 68 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online