Re: math problems
rockerpham wrote:I have two math problems that need your help,
1. A machine part is manufactured at a unit cost of $0.40 for materials and $0.15 for direct labor. An investment of $500,000 in tooling is required. The order calls for 3 million pieces.
Halfway through the order, a new method of manufacture can be put into effect that will reduce the unit costs to $0.35 for material and $0.14 for direct labor?-but will require $100,000 for additional tooling. This tooling will not be useful for future orders.
Other costs are allocated at 2.5 times the direct labor costs.
Complete an analysis.
cost1 = 0.55*3e6 + 5e5 = 2.15e6
cost2 = 0.55*1.5e6 + 5e5 + 0.49*1.5e6 + 1e5 = 2.16e6
Sounds like option 2, ignoring "other costs", will save $10,000. Future orders are irrelevant.
Quote:2. In the design of a cold-storage warehouse, the specifications call for a maximum heat transfer through the warehouse walls of 30,000 joules per hour per square meter (30,000 joules/hr/m2) when there is a 30oC temperature difference between the inside surface and the outside surface of the insulation.
The two insulation materials being considered are as follows:
Insulation Material Cost per Cubic Meter Conductivity
(J-m/m2-oC-hour)
Rock wool $12.00 125
Foamed insulation $15.00 100
The basic equation for heat conduction through a wall is:
Q = (K(ΔT))/L
Where:
Q = heat transfer, in J/hr/m2 of wall.
K = conductivity in J-m/m2-oC-hour.
ΔT = temperature difference between the two surfaces in oC.
L = thickness of insulating material, in meters.
What are the total costs of each insulation material?
Heat transfer:
3e5 = K*30/L
Costs per square meter:
cost1 = 12*L1 = 12*125*30/3e5 = 0.15
cost2 = 15*L2 = 15*100*30/3e5 = 0.15
Same cost dude. Might as well go with the foamed insulation since it takes up less space and is easy to install.
/Kayyam