Welcome :: Homework Help and Answers :: Mathskey.com

Recent Visits

Calculus, 8th Edition,stewart; page 22 problem 63

A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 12 in. by 20 in. by cutting out equal squares of side x at each corner and then folding up the sides as in the figure. Express the volume V of the box as a function of x.

asked Jul 28, 2015 in CALCULUS by anonymous

Share this question

1 Answer

Step:1

The dimensions of the cardboard $12$ in. by $20$ in.

Observe the figure:

The height of the box is $x$ in.

Width of the box = width of the cardboard $-2x=12-2x$ .

Length of the box = length of the cardboard $-2x=20-2x$ .

Step:2

Volume of the box = Length x Width x Height.

Volume of the box $=(20-2x)(12-2x)(x)$

$=(240-40x-24x+4x^2)(x)$

$=x(240-64x+4x^2)$

$=240x-64x^2+4x^3$

$=4x^3-64x^2+240x$

The volume of the box as a function of $x$ is $4x^3-64x^2+240x$ .

Step:3

Domain :

All possible values of $x$ is the domain of the function.

Volume of the box should be greater than zero.

So $L>0$ , $W>0$ and $H>0$ .

$20-2x>0$ , $12-2x>0$ and $x>0$ .

$x<10$ , $x<6$ and $x>0$ .

The domain of the function is $0<x<6$ .

Solution:

The volume of the box as a function of $x$ is $4x^3-64x^2+240x$ , $0<x<6$ .

answered Jul 28, 2015 by dozey Mentor

...