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Chapter 10: Problem 20

A manufacturer wants to procure rectangular boxes to ship its product. The boxes must contain 20 cubic feet of space. To be durable enough to ensure the safety of the product, the material for the sides of the boxes will cost \(\$ 0.10\) per square foot, while the material for the top and bottom will cost \(\$ 0.25\) per square foot. In this activity we will help the manufacturer determine the box of minimal cost. a. What quantities are constant in this problem? What are the variables in this problem? Provide appropriate variable labels. What, if any, restrictions are there on the variables? b. Using your variables from (a), determine a formula for the total cost \(C\) of a box. c. Your formula in part (b) might be in terms of three variables. If so, find a relationship between the variables, and then use this relationship to write \(C\) as a function of only two independent variables. d. Find the dimensions that minimize the cost of a box. Be sure to verify that you have a minimum cost.

Short Answer

Expert verified
The dimensions that minimize the cost of a box are: Length: \(l = \sqrt[3]{50}\), Width: \(w = \sqrt[3]{40}\), and Height: \(h = \sqrt[3]{\frac{1}{100}}\). These dimensions will result in the minimum cost for the manufacturer.

Step by step solution

01

Identifying Constants and Variables

Constants: - \(20\) cubic feet (Volume of the box) - \(0.10\) per square foot (Cost of the material for the sides) - \(0.25\) per square foot (Cost of the material for the top and bottom) Variables: - Length, width, and height of the box. We can use \(l\), \(w\), and \(h\) to represent them, respectively. Restrictions: All dimensions (\(l\), \(w\), and \(h\)) must be positive for a meaningful box.
02

Determining a Formula for the Total Cost

We can calculate the cost of each side of the box and then add them up to get the total cost. Cost of the sides (2 lw sides and 2 wh sides): \(2lw(0.10)+2wh(0.10) = 0.20lw + 0.20wh\) Cost of the top and bottom (2 lh sides): \(2lh(0.25) = 0.50lh\) The total cost \(C\) is the sum of the cost of all sides: \(C = 0.20lw + 0.20wh + 0.50lh\)
03

Reducing the Number of Variables

Since the volume of the box is fixed, we have: \(V = lwh = 20\) We can solve for one of the variables, for example, \(h\): \(h = \frac{20}{lw}\) Now, we can substitute this into the cost function to eliminate \(h\) and express \(C\) as a function of the two independent variables \(l\) and \(w\). \(C(l, w) = 0.20lw + 0.20w\frac{20}{lw} + 0.50l\frac{20}{lw} = 0.20lw + 4w + 10l\)
04

Finding the Dimensions that Minimize the Cost

To find the dimensions that minimize the cost, we need to find the critical points of the cost function \(C(l, w)\) and check if it is a minimum. Now, let's find the partial derivatives of \(C\) with respect to \(l\) and \(w\), and set them equal to zero: \(\frac{\partial C}{\partial l} = 0.20w - 10\frac{1}{w} = 0\) \(\frac{\partial C}{\partial w} = 0.20l - 4\frac{1}{l} = 0\) Solving the system of equations, we get: \(l = \sqrt[3]{50}\) \(w = \sqrt[3]{40}\) To verify that this critical point corresponds to a minimum, we can use the second derivative test. We need to find the second order partial derivatives and compute the Hessian determinant: \(\frac{\partial^2 C}{\partial l^2} = 10\frac{1}{w^2} > 0\) \(\frac{\partial^2 C}{\partial w^2} = 4\frac{1}{l^2} > 0\) \(\frac{\partial^2 C}{\partial l \partial w} = 0\) Hessian determinant: \(D =\frac{\partial^2 C}{\partial l^2}\frac{\partial^2 C}{\partial w^2} -(\frac{\partial^2 C}{\partial l \partial w})^2 = 40\frac{1}{l^2w^2} > 0\) Since the Hessian determinant is positive and the second partial derivative with respect to \(l\) is also positive, this critical point corresponds to a minimum. Therefore, the dimensions that minimize the cost of a box are: Length: \(l = \sqrt[3]{50}\) Width: \(w = \sqrt[3]{40}\) Height: \(h = \frac{20}{lw} = \frac{20}{(\sqrt[3]{50})(\sqrt[3]{40})} = \sqrt[3]{\frac{1}{100}}\) These dimensions will result in the minimum cost for the manufacturer.

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Key Concepts

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

Cost Minimization
Cost minimization is a standard application of calculus where a company wants to produce goods while keeping its expenditure as low as possible. To achieve this, they need to determine the combination of factors鈥攕uch as dimensions of a product, material used, quantity produced鈥攖hat result in the least possible cost while still meeting all the required specifications. In the given problem, a manufacturer needs to design boxes with a specific volume that is made from materials of different costs per unit area.

The mathematical model used for cost minimization will typically involve creating a cost function that includes all necessary variables and constraints. The process involves calculating each component's cost and summing these to obtain the total cost. In this example, the sides, top, and bottom of the box have different costs, leading to a formula involving the length, width, and height of the box, linked by their relationship to the box's volume.
Multivariable Calculus
Multivariable calculus extends single variable calculus concepts to functions of several variables. It involves partial derivatives, gradients, and multiple integrals. In problems like our box design, we use multivariable calculus to deal with more than one independent variable鈥攊n this case, the length and width of the box.

Our objective is to express the cost function in terms of these variables while considering the fixed volume of the box as a constraint. This example illustrates how real-world problems often depend on several factors and require optimization across multiple variables to find the best solution.
Partial Derivatives
Partial derivatives are a crucial tool in multivariable calculus used to measure how a function changes as each variable is varied while keeping other variables constant. In the context of our optimization problem, we calculate the partial derivatives of the cost function with respect to each of the box's dimensions.

This reveals the relationship between a slight change in dimension and its impact on cost. When we set these partial derivatives to zero, we find the critical points of the function, which indicate where a local minimum or maximum may exist. The process involves differentiating with respect to one variable at a time, while treating the other variables as constants. This incremental approach helps in managing the complexity of functions that depend on multiple variables.
Applications of Derivatives
Derivatives are not just theoretical constructs; they have practical applications in a wide range of fields, including economics, engineering, and physical sciences. The manufacturer's problem is an example of how derivatives guide us towards a minimum cost in production.

By applying the first and second derivative tests, we verify whether the critical points found by setting the first derivatives to zero represent a minimum cost situation. The use of the second derivative鈥攐r in this case, the second order partial derivatives鈥攈elps confirm the nature of the critical points and ensures the solution will indeed minimize costs.

This process exemplifies the pivotal role that derivatives play in making informed decisions in a multitude of real-world optimization scenarios.

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Most popular questions from this chapter

One mole of ammonia gas is contained in a vessel which is capable of changing its volume (a compartment sealed by a piston, for example). The total energy \(U\) (in Joules) of the ammonia is a function of the volume \(V\) (in cubic meters) of the container, and the temperature \(T\) (in degrees Kelvin) of the gas. The differential \(d U\) is given by \(d U=840 d V+27.32 d T\). (a) How does the energy change if the volume is held constant and the temperature is decreased slightly? \(\odot\) it increases slightly \(\odot\) it does not change \(\odot\) it decreases slightly (b) How does the energy change if the temperature is held constant and the volume is increased slightly? \(\odot\) it does not change \(\odot\) it increases slightly \(\odot\) it decreases slightly (c) Find the approximate change in energy if the gas is compressed by 150 cubic centimeters and heated by 3 degrees Kelvin. Change in energy = ___________. Please include units in your answer.

Find all the first and second order partial derivatives of \(f(x, y)=3 \sin (2 x+\) \(y)-4 \cos (x-y)\) A. \(\frac{\partial f}{\partial x}=f_{x}=\) ________. B. \(\frac{\partial f}{\partial y}=f_{y}=\) ________. C. \(\frac{\partial^{2} f}{\partial x^{2}}=f_{x x}=\) ________. D. \(\frac{\partial^{2} f}{\partial y^{2}}=f_{y y}=\) ________. E. \(\frac{\partial^{2} f}{\partial x \partial y}=f_{y x}=\) ________. F. \(\frac{\partial^{2} f}{\partial y \partial x}=f_{x y}=\) ________.

The concentration of salt in a fluid at \((x, y, z)\) is given by \(F(x, y, z)=\) \(2 x^{2}+3 y^{4}+2 x^{2} z^{2} \mathrm{mg} / \mathrm{cm}^{3}\). You are at the point (-1,1,-1) . (a) In which direction should you move if you want the concentration to increase the fastest? direction: _________. (Give your answer as a vector.) (b) You start to move in the direction you found in part (a) at a speed of \(5 \mathrm{~cm} / \mathrm{sec} .\) How fast is the concentration changing? rate of change \(=\) ______________.

(a) Check the local linearity of \(f(x, y)=e^{-x} \cos (y)\) near \(x=1, y=1.5\) by filling in the following table of values of \(f\) for \(x=0.9,1,1.1\) and \(y=1.4,1.5,1.6 .\) Express values of \(f\) with 4 digits after the decimal point. (b) Next, fill in the table for the values \(x=0.99,1,1.01\) and \(y=\) 1.49,1.5,1.51 , again showing 4 digits after the decimal point. Notice if the two tables look nearly linear, and whether the second looks more linear than the first (in particular, think about how you would decide if they were linear, or if the one were more closely linear than the other). (c) Give the local linearization of \(f(x, y)=e^{-x} \cos (y)\) at (1,1.5) : Using the second of your tables: \(f(x, y) \approx\) ____________. Using the fact that \(f_{x}(x, y)=-e^{-x} \cos (y)\) and \(f_{y}(x, y)=-e^{-x} \sin (y):\) \(f(x, y) \approx\) ___________.

Suppose that you are climbing a hill whose shape is given by \(z=902-\) \(0.07 x^{2}-0.1 y^{2},\) and that you are at the point (40,70,300) In which direction (unit vector) should you proceed initially in order to reach the top of the hill fastest? (_________________) If you climb in that direction, at what angle above the horizontal will you be climbing initially (radian measure)?
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