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Chapter 4: Problem 7

Optimal Dimensions You are planning to make an open rectangular box from an 8- by 15-in. piece of cardboard by cutting congruent squares from the corners and folding up the sides. What are the dimensions of the box of largest volume you can make this way, and what is its volume?

Short Answer

Expert verified
The dimensions of the box with the largest volume are 11 in. by 4 in. by 2 in. and the largest volume is 88 cubic inches.

Step by step solution

01

Define Variables

A square will be cut from each corner of the cardboard. Let's denote the side of this square as \(x\) (in inches). Therefore, the dimensions of the box will be (15 - 2x) in length, (8 - 2x) in width, and \(x\) in height.
02

Formulate the Volume function

The volume \(V\) of a box is given by the formula \(V = length \times width \times height\). Plugging the dimensions obtained in the previous step into the volume formula gives us the function \(V(x) = x(15 - 2x)(8 - 2x)\).
03

Find the domain of \(V\)

The domain of \(V\) depends on the physical dimensions of the cardboard, meaning that \(x\) must be greater than 0 and less than or equal to the smallest half-dimension of the cardboard, which is 4. Therefore, the domain of \(V\) is (0,4).
04

Differentiate the volume function

To find the maximum volume, differentiate the function \(V(x) = x(15 - 2x)(8 - 2x)\) with respect to \(x\), which gives \(V'(x) = -4x^2 + 46x - 120\).
05

Solve for critical points

Set \(V'(x)\) equal to zero and solve for \(x\) to find the critical points. Solving \(-4x^2 + 46x - 120 = 0\) yields, \(x = 2\) and \(x = 15\).
06

Test critical points within the domain

Since \(x = 15\) is not in the domain of \(V\), reject it. Therefore, \(x = 2\) in. is the dimension of the squares to be cut from the corners. Thus, the dimensions of the box are Length: 15 - 2*2 = 11 in., Width: 8 - 2*2 = 4 in., and Height: 2 in.
07

Calculate the maximum volume

Substitute the dimensions into the volume formula to get the maximum volume: \(V = Length \times Width \times Height = 11 \times 4 \times 2 = 88\) cubic inches.

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Most popular questions from this chapter

In Exercises 23 and \(24,\) a particle is moving along the curve \(y=f(x) .\) \(y=f(x)=\frac{10}{1+x^{2}}\) If \( \)d x / d t=3 \mathrm{cm} / \mathrm{sec}, \text { find } d y / d t \(d x / d t=3 \mathrm{cm} / \mathrm{sec},\) find \(d y / d t\) at the point where $$x=-2 \text { . } \quad \text { (b) } x=0 . \quad \text { (c) } x=20$$

Multiple Choice What is the linearization of \(f(x)=e^{x}\) at \(x=1 ?\) (A) \(y=e \quad\) (B) \(y=e x \quad\) (C) \(y=e^{x}\) \((\mathbf{D}) y=x-e \quad\) (\mathbf{E} ) ~ \(y=e(x-1)\)

Finding Parameter Values Show that \(f(x)=x^{2}+(a / x)\) cannot have a local maximum for any value of \(a .\)

Stiffness of a Beam The stiffness S of a rectangular beam is proportional to its width times the cube of its depth. (a) Find the dimensions of the stiffest beam that can be cut from a 12-in. diameter cylindrical log. (b) Writing to Learn Graph \(S\) as a function of the beam's width \(w,\) assuming the proportionality constant to be \(k=1 .\) Reconcile what you see with your answer in part (a). (c) Writing to Learn On the same screen, graph \(S\) as a function of the beam's depth \(d,\) again taking \(k=1 .\) Compare the graphs with one another and with your answer in part (a). What would be the effect of changing to some other value of \(k ?\) Try it.

Estimating Volume You can estimate the volume of a sphere by measuring its circumference with a tape measure, dividing by 2\(\pi\) to get the radius, then using the radius in the volume formula. Find how sensitive your volume estimate is to a 1\(/ 8\) in. error in the circumference measurement by filling in the table below for spheres of the given sizes. Use differentials when filling in the last column. $$\begin{array}{|c|c|c|}\hline \text { Sphere Type } & {\text { True Radius }} & {\text { Tape Error Radius Error Volume Error }} \\ \hline \text { Orange } & {2 \text { in. }} & {1 / 8 \text { in. }} \\ \hline \text { Melon } & {4 \text { in. }} & {1 / 8 \text { in. }} \\ \hline \text { Beach Ball } & {7 \text { in. }} & {1 / 8 \text { in. }}\end{array}$$
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