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In Exercises \(35-40\) , find the partial derivative of the function with respect to each variable. Wilson lot size formula (Section \(4.5,\) Exercise 53 ) $$A(c, h, k, m, q)=\frac{k m}{q}+c m+\frac{h q}{2}$$

Short Answer

Expert verified
Partial derivatives are: \(m, \frac{q}{2}, \frac{m}{q}, \frac{k}{q} + c, -\frac{k m}{q^2} + \frac{h}{2}\).

Step by step solution

01

Understanding the Problem Statement

We need to find the partial derivatives of the given function \(A(c, h, k, m, q)\) with respect to all its variables: \(c\), \(h\), \(k\), \(m\), and \(q\). This means treating all other variables as constants while differentiating with respect to one variable at a time.
02

Partial Derivative with Respect to \(c\)

To find \(\frac{\partial A}{\partial c}\), differentiate \(A\) treating \(c\) as the variable and the others as constants. The term involving \(c\) is \(c m\). Thus, \(\frac{\partial A}{\partial c} = m\).
03

Partial Derivative with Respect to \(h\)

For the derivative \(\frac{\partial A}{\partial h}\), the only term with \(h\) is \(\frac{h q}{2}\). Differentiating this term with respect to \(h\), we get \(\frac{q}{2}\). Thus, \(\frac{\partial A}{\partial h} = \frac{q}{2}\).
04

Partial Derivative with Respect to \(k\)

To find \(\frac{\partial A}{\partial k}\), we consider the term \(\frac{k m}{q}\). The derivative of this term with respect to \(k\) is \(\frac{m}{q}\), making \(\frac{\partial A}{\partial k} = \frac{m}{q}\).
05

Partial Derivative with Respect to \(m\)

Differentiating \(A\) with respect to \(m\), we focus on \(\frac{k m}{q} + c m\). The derivative of these terms gives \(\frac{k}{q} + c\), so \(\frac{\partial A}{\partial m} = \frac{k}{q} + c\).
06

Partial Derivative with Respect to \(q\)

For \(\frac{\partial A}{\partial q}\), differentiate \(\frac{k m}{q} + \frac{h q}{2}\). The derivative is \(-\frac{k m}{q^2} + \frac{h}{2}\), so \(\frac{\partial A}{\partial q} = -\frac{k m}{q^2} + \frac{h}{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wilson lot size formula
The Wilson lot size formula is a well-known equation used in inventory management. It helps in determining the optimal order quantity that minimizes the total cost associated with ordering and holding inventory. The formula balances order costs and holding costs efficiently to avoid excess stock while ensuring supplies are available.

The basic idea behind the Wilson formula can be summed up in a few concepts:
  • **Ordering Costs:** These are the costs involved every time an order is placed. The costs include the processing, shipping, and handling of goods.
  • **Holding Costs:** These costs occur from storing unsold goods. It includes warehousing expenses, depreciation, and opportunity costs of tied-up capital.
Applying the Wilson lot size formula allows businesses to calculate the ideal order size by considering these two costs as variables. The application in multivariable calculus, like the problem given, aims to identify how changes in one variable, such as unit cost or holding costs, affect the minimum cost order quantity. By using calculus, companies can effectively manage their inventories to optimize costs.
Multivariable Calculus
Multivariable calculus extends the concepts of single-variable calculus to functions of several variables. It is particularly useful when dealing with equations that depend on more than one variable, such as the function in the given exercise, which depends on five variables: \(c\), \(h\), \(k\), \(m\), and \(q\).

This branch of calculus allows you to explore how each independent variable affects the outcome of the function. It's essential in fields like economics, physics, and engineering, where systems often have multiple varying factors.

Key concepts in multivariable calculus include:
  • **Partial Derivatives:** These show the rate of change of a function with respect to one variable while keeping others constant, as seen in the exercise.
  • **Gradient:** A vector that comprises all the partial derivatives of a function. It's used to determine the direction of steepest ascent or descent.
Understanding these principles provides a robust toolset for analyzing and solving complex real-world problems with multiple variables.
Differentiation
Differentiation is a fundamental concept in calculus, and it involves finding the derivative of a function. The derivative measures how a function changes with respect to changing inputs. In single-variable calculus, this is pretty straightforward, but when you get into multiple variables, as with the given exercise, it involves finding partial derivatives.

Let's break down the process of differentiation with respect to multiple variables:
  • **Single Variable:** Here, the derivative is found with respect to one function's variable, showing how the function changes as that variable changes.
  • **Multiple Variables (Partial Derivatives):** When a function has several variables, a partial derivative gives the function's rate of change concerning one variable while treating the others as constants.
For instance, in the given problem, determining the partial derivative of \(A\) with respect to any of its variables like \(c\), \(h\), \(k\), \(m\), and \(q\) involves focusing on terms containing that specific variable only, treating other parameters as if they were constant in that context. This approach is crucial for understanding how changing one component while holding others constant can affect a system or function comprehensively.

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Most popular questions from this chapter

Variation in electrical resistance by wiring resistors of \(R_{1}\) and \(R_{2}\) ohms in parallel (see accompanying figure) can be calculated from the formula $$ \frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}} $$ a. Show that $$ d R=\left(\frac{R}{R_{1}}\right)^{2} d R_{1}+\left(\frac{R}{R_{2}}\right)^{2} d R_{2} $$ b. You have designed a two-resistor circuit, like the one shown, to have resistances of \(R_{1}=100\) ohms and \(R_{2}=400\) ohms, but there is always some variation in manufacturing and the resistors received by your firm will probably not have these exact values. Will the value of \(R\) be more sensitive to variation in \(R_{1}\) or to variation in \(R_{2} ?\) Give reasons for your answer. c. In another circuit like the one shown, you plan to change \(R_{1}\) from 20 to 20.1 ohms and \(R_{2}\) from 25 to 24.9 ohms. By about what percentage will this change \(R ?\)

You will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$f(x, y)=\left\\{\begin{array}{ll}{x^{5} \ln \left(x^{2}+y^{2}\right),} & {(x, y) \neq(0,0)} \\ {0,} & {(x, y)=(0,0)}\end{array}\right.$$ $$-2 \leq x \leq 2, \quad-2 \leq y \leq 2$$

In Exercises \(51-54,\) verify that \(w_{x y}=w_{y x}\) $$w=x y^{2}+x^{2} y^{3}+x^{3} y^{4}$$

A least squares plane The plane \(z=A x+B y+C\) is to be "fitted" to the following points \(\left(x_{k}, y_{k}, z_{k}\right) :\) $$(0,0,0), \quad(0,1,1), \quad(1,1,1), \quad(1,0,-1).$$ Find the values of \(A, B,\) and \(C\) that minimize $$\sum_{k=1}^{4}\left(A x_{k}+B y_{k}+C-z_{k}\right)^{2},$$ the sum of the squares of the deviations.

Find the linearization \(L(x, y, z)\) of the function \(f(x, y, z)\) at \(P_{0} .\) Then find an upper bound for the magnitude of the error \(E\) in the approximation \(f(x, y, z) \approx L(x, y, z)\) over the region \(R\) $$ \begin{array}{l}{f(x, y, z)=x z-3 y z+2 \text { at } P_{0}(1,1,2)} \\ {R :|x-1| \leq 0.01, \quad|y-1| \leq 0.01, \quad|z-2| \leq 0.02}\end{array} $$

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