A manufacturer of hospital supplies has a uniform annual demand for 80,000 boxes of
bandages. It costs $10 to store one box of bandages for one year and $160 to set up
the plant for prduction. Haw many times a year should the company produce boxes
of bandages in order to minimize the total storage and setup costs?
To determine the optimal production frequency, we need to minimize the total storage and setup costs.
Let's break down the costs involved:
1. Storage costs: It costs $10 to store one box of bandages for one year. So, for 80,000 boxes, the storage cost per year would be 80,000 * $10.
2. Setup costs: It costs $160 to set up the plant for production. These setup costs are incurred each time production is initiated.
3. Production frequency: We need to find out how many times a year the company should produce boxes of bandages.
Now, let's assume the company produces 'x' times a year. In this case, each production batch should consist of 80,000 / x boxes.
The total cost equation can be written as:
Total Cost = (Storage Cost per year) + (Number of setups per year * Setup Cost)
The storage cost per year is given by:
Storage Cost per year = (Number of boxes in each batch * Storage cost per box) * (Number of batches per year)
Storage Cost per year = ((80,000 / x) * $10) * x
Storage Cost per year = $800,000
The number of setups per year is given by:
Number of setups per year = Number of times production is initiated
Number of setups per year = x
The total cost equation becomes:
Total Cost = $800,000 + (x * $160)
To minimize the total cost, we can take the derivative of the total cost with respect to 'x' and set it equal to zero:
d(Total Cost) / dx = 0
d($800,000 + (x * $160)) / dx = 0
$160 = 0
Since the derivative is a constant, it equals zero. We can ignore it.
Now, plug in the value of x into the equation:
Number of setups per year = x
Number of setups per year = Number of times production is initiated
Number of setups per year = x
To minimize the total cost, the company should produce boxes of bandages once a year, resulting in the lowest total cost.