   Chapter 10.4, Problem 26E ### Mathematical Applications for the ...

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

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

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

# Inventory cost model A company needs 450,000 items per year. Production costs are $500 to prepare for a production run and$10 for each item produced. Inventory costs are $2 per item per year. Find the number of items that should be produced in each run so that the total costs of production and storage are minimized. 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$500. The cost of producing one item is $10 and that of storage is$2 per year. Items required are 450,000.

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 $500. The cost of producing one item is$10 and that of storage is \$2 per year. Items required are 450,000.

Take number of items x in one run. The cost is C=450,000x(500)+450,000(10)+(x2)2 as x2 is the average units in storage

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