- #1
space-time
- 218
- 4
- Homework Statement
- The demand for some bikes that a company sells is 10000 bikes per year, and they sell at a uniform rate throughout the year. The cost of one shipment of bikes is $10000 and the cost of storing each bike is $200 per year. If the company orders too many bikes at once, the storage cost increases, and if they order too often, the shipment cost increases. How large should each order be, and how often should the orders be placed to minimize the ordering and storage costs?
- Relevant Equations
- N/A (or at least I don't know of any specific equations that are particularly relevant here)
I first tried to set up an expression for the total amount of money that the company would spend in a year. That would be:
p(x, y) = 10000x + 200xy
where:
p(x, y) = the amount of money (dollars) that the company would spend in a year.
x = the number of shipments ordered per year.
y = the number of bikes per shipment
Now, I was thinking that this would be an optimization problem where I would have to find a local minimum for the function p. However,
px = 10000 + 200y
py = 200x
These partial derivatives only ever equal 0 at x = 0 and y = -50.
That presents quite the problem, seeing as how the company would certainly need to order more than 0 shipments per year, and there cannot be -50 bikes per shipment.
Hence, I am stuck. Is my approach all wrong to begin with?
p(x, y) = 10000x + 200xy
where:
p(x, y) = the amount of money (dollars) that the company would spend in a year.
x = the number of shipments ordered per year.
y = the number of bikes per shipment
Now, I was thinking that this would be an optimization problem where I would have to find a local minimum for the function p. However,
px = 10000 + 200y
py = 200x
These partial derivatives only ever equal 0 at x = 0 and y = -50.
That presents quite the problem, seeing as how the company would certainly need to order more than 0 shipments per year, and there cannot be -50 bikes per shipment.
Hence, I am stuck. Is my approach all wrong to begin with?