Chapter 4.8, Problem 52E

(a)

To determine

To calculate: The length of the cuts to make the rectangular box.

Explanation of Solution

Given information:

Length of the cardboard box $=20\text{inch}$

Breadth of the cardboard box $=16\text{inch}$

Calculation:

Let the height of the rectangular box be $x$

If the sides of the cardboard are cut then, length of the rectangular box $=20-2x$  ……. (1)

Breadth of the rectangular box $=16-2x$  …… (2)

(b)

To determine

To calculate: The maximum volume of the rectangular box

Explanation of Solution

Given information:

Length of the cardboard box $=20\text{inch}$

Breadth of the cardboard box $=16\text{inch}$

Formula used:

Volume=length × breadth × height

Calculation:

Let the height of the rectangular box be $x$

If the sides of the cardboard are cut then, length of the rectangular box $=20-2x$

Breadth of the rectangular box $=16-2x$

  $\begin{array}{c}V(x)=(20-2x)(16-2x)x\\ =320x-40x^2-32x^2+4x^3\\ =4x^3-72x^2+320x\end{array}$

By graphing the function we get

  

Hence, the maximum volume is $420.11{\text{inch}}^{\text{3}}$ for maximum value of $x=2.945\text{inch}$

(c)

To determine

To calculate: The dimensions of the rectangular box.

Explanation of Solution

Given information:

Length of the cardboard box $=20\text{inch}$

Breadth of the cardboard box $=16\text{inch}$

Formula used:

Length of the rectangular box $=20-2x$

Breadth of the rectangular box $=16-2x$

Calculation:

Let the height of the rectangular box be $x$

If the sides of the cardboard are cut then, length of the rectangular box $=20-2x$

Breadth of the rectangular box $=16-2x$

  $\begin{array}{c}V(x)=(20-2x)(16-2x)x\\ =320x-40x^2-32x^2+4x^3\\ =4x^3-72x^2+320x\end{array}$

By graphing the function we get

  

Hence, the maximum volume is $420.11{\text{inch}}^{\text{3}}$ for maximum value of $x=2.945\text{inch}$

So, the length of the box is $=20-2x=20-2\times 2.95=14.1\text{inch}$

The breadth of the box is $=16-2x=16-2\times 2.95=10.1\text{inch}$