# World of Functions

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• Jul 20th 2008, 07:24 AM
magentarita
World of Functions
A open box with a square base is required to have a volume of 10 cubic feet.

(a) Express the amount A of material used to make such a box as a function of the length x of a side of the square base.

(b) How much material is required for a base 1 foot by 1 foot?
• Jul 20th 2008, 11:46 PM
earboth
Quote:

Originally Posted by magentarita
A open box with a square base is required to have a volume of 10 cubic feet.

(a) Express the amount A of material used to make such a box as a function of the length x of a side of the square base.

(b) How much material is required for a base 1 foot by 1 foot?

Let h denote the height of the box. Then the volume of the box is:

$V = x^2 \cdot h = 10\ cft~\implies~ \boxed{h = \frac{10}{x^2}}$

The surface area of the box exists of 1 square and 4 congruent rectangles:

$A = x^2+4 \cdot h \cdot x$ Plug in the term for h from the first equation to eliminate h:

$A(x)=x^2+4 \cdot \frac{10}{x^2} \cdot x = x^2+\frac{40}x$

To calculate the extreme amount of material use the first derivation of A:

$A'(x) = 2x-\frac{40}{x^2}$ . Solve the equation $A'(x) = 0$ for x. For your confirmation only: $x = \sqrt[3]{20} \approx 2.71\ ft$
• Jul 21st 2008, 04:53 AM
magentarita
Your x^2
Where did x^2 come from?

Also, isn't the volume of a box V = length times width times height?

Where is this formula in your reply?

Thanks,
Rita
• Jul 21st 2008, 05:15 AM
earboth
Quote:

Originally Posted by magentarita
Where did x^2 come from?

Also, isn't the volume of a box V = length times width times height?

Where is this formula in your reply?

Thanks,
Rita

Your problem said: "A open box with a square base is ..."

That means the length and the width are equal. Since one side of the square should have been labeled by x the area of the base is $x^2$ and the volume of the box is calculated by:

$volume = (base\ area) \cdot (height)$ or using one letter variable names:

$v = x^2 \cdot h$
• Jul 21st 2008, 06:38 AM
magentarita
Got It!
I totally get it now.

Good job! Thank you.

This is a great site! I can meet great math teachers here and learn enough math to pass my course.

Thank you very much.
• Oct 29th 2012, 12:52 AM
studentofmath89
Re: World of Functions
so how do you express the ammount of material that is needed as a function of the length x. if u allready explained please elaborate
• Oct 29th 2012, 01:02 AM
MarkFL
Re: World of Functions
We have:

$x^2h=V$ where $V$ is a given constant. Hence:

$h=\frac{V}{x^2}$

The amount of material $A$ is the area of the bottom and 4 sides:

$A=x^2+4xh$ and so:

$A(x)=x^2+4x\left(\frac{V}{x^2} \right)=x^2+\frac{4V}{x}$