# Volume of a box?

• Sep 28th 2008, 06:09 PM
realintegerz
Volume of a box?
Find the volume of a box as a function of the width of base of the box?

If 12,000 cm^2 of material is available to make a box with a square base and an open top, find the volume of the box as a function of the width if the base of the box
• Sep 28th 2008, 07:04 PM
Jhevon
Quote:

Originally Posted by realintegerz
Find the volume of a box as a function of the width of base of the box?

If 12,000 cm^2 of material is available to make a box with a square base and an open top, find the volume of the box as a function of the width if the base of the box

did you draw a diagram? (see the attached).

let the width of the base be $x$ and the height of the box be $y$

thus the volume (length times width times height) is given by

$V = x^2 y$

Now we know the surface area is 12000, this is the sum of the areas of all the faces, so

$12000 = \underbrace{x^2}_{\text{area of base}} + \underbrace{4xy}_{\text{area of the 4 sides}}$

that is $12000 = x^2 + 4xy$

now can you answer the problem?
• Sep 28th 2008, 07:13 PM
Soroban
Hello, realintegerz!

Quote:

If 12,000 cm² of material is available to make a box
with a square base and an open top, find the volume of the box
as a function of the width of the base of the box.

Code:

        *-------*         /|      /|       / |    / |       *-------*  |       |      |  |       |      |  |     h|      |  *       |      | /       |      |/ x       *-------*           x

Let $x$ = side of the square base.
Let $h$ = height of the box.

The panels comprising the box is made up of:
. . a base with area $x^2$
. . and four sides with area $xh$ each.
The total surface area is: . $x^2 + 4xh$

We are told that the total surface area will be 12,000 cm².

So we have: . $x^2 + 4xh \:=\:12,\!000 \quad\Rightarrow\quad h \:=\:\frac{12,\!000-x^2}{4x}$ .[1]

The volume of the box is: . $V \;=\;x^2h$

Substitute [1]: . $V \;=\;x^2\left(\frac{12,\!000-x^2}{4x}\right) \quad\Rightarrow\quad \boxed{V \;=\;\frac{1}{4}(12,\!000x - x^3)}$