Chapter 10.4, Problem 24E

### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

Chapter
Section

### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Inventory cost model Suppose that a company needs 60,000 items during a year and that preparation for each production run costs $400. Suppose further that it costs$4 to produce each item and $0.75 to store an item for one year. Use the inventory cost model to find the number of items in each production run that will minimize the total costs of production and storage. To determine To calculate: The number of items in one run to minimize the cost of production and storage. Explanation Given Information: The preparation for production costs$400. The cost of producing one item is $4 and that of storage is$0.75 per year. Items required are 60,000.

Formula used:

To find the minimum value, calculate the relevant stationary value of an equation. Differentiate the function with respect to the independent variable and equate it to 0. the relevant value is the stationary value that satisfies the provided conditions.

If the second derivative of the provided function is more than zero, then substituting the value of the independent variable will give the minimum value of the equation.

The power rule is used for a function in which the expression can be written as every term raised to a power (be it fractional, positive or negative). For the function f(x)=xn, the derivative is

ddx[xn]=nxn1

Calculation:

Consider the provided statement,

The preparation for production costs $400. The cost of producing one item is$4 and that of storage is \$0.75 per year. Items required are 60,000.

Take number of items x in one run. The cost is C=60,000x(400)+60,000(4)+(x2)0.75 as x2 is the average units in storage.

To find the stationary value for C=60,000x(400)+60,000(4)+(x2)0.75, differentiate the equation with respect to the independent variable x to get

C'=60,000×400x2+0

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