Chapter 1.4, Problem 61E

(a)

To determine

Determine the maximum volume from the given table.

Answer to Problem 61E

The maximum volume from the given table is $V=1024$ .

Explanation of Solution

Given:

Consider an open box of maximum volume (in form of cuboid) with following dimensions.

Length is $24–2x$

Breath is $24–2x$

Height is x

Also,

Height, x	1	2	3	4	5	6
Volume, V	484	800	972	1024	980	864

Let estimate the maximum volume from the table.

From the table it is observe that height with $x=4$ has maximum volume $V=1024$ .

(b)

To determine

Plot the graph from the table obtained in part (a).

Answer to Problem 61E

Graph is shown below.

Explanation of Solution

Given:

Consider an open box of maximum volume (in form of cuboid) with following dimensions.

Length is $24–2x$

Breath is $24–2x$

Height is x

Let plot the graph for the points height and volume $\left(x,V\right)$

  

Graphically, each value of height, $x$ is matched with exactly one value of volume, $V$ . Thus, $V$ is as function of $x$ .

(c)

To determine

Find the function and determine the domain.

Answer to Problem 61E

The function is :

  $V=x{\left(24–2x\right)}^2$

The domain is $0<x<12$

Explanation of Solution

Given:

V is a function of x

V is a function of x.

Let the function of V in terms of x.

Volume of cuboid is length × breadth × height

  $\begin{array}{l}V=\left(24–2x\right)\times \left(24–2x\right)\times \left(x\right)\hfill \\ =x{\left(24–2x\right)}^2\hfill \end{array}$

Hence, V is as function of x.

Now find the domain of the function. Since height and volume cannot be negative or 0.

Therefore,

  $x{\left(24–2x\right)}^2\ge 0$

Then $x>0$ or

  $\begin{array}{l}{\left(24–2x\right)}^2\ge 0\hfill \\ \left(24–2x\right)>0\hfill \\ 24>2x\hfill \\ x<\frac{24}{2}\hfill \\ x<12\hfill \end{array}$

Hence, the domain is $0<x<12$