# Prove surface area of cuboid.

• Dec 27th 2009, 01:35 AM
BabyMilo
Prove surface area of cuboid.
Quote:

Quote:

A cuboid has a volume of $8 m^3$. The base of the cuboid is square with sides of length x metres. the sureface area of the cuboid is $A m^2$.
Quote:

Show that $A=2x^2+\frac{32}{x}$

thanks!
• Dec 27th 2009, 02:15 AM
Prove It
Quote:

Originally Posted by BabyMilo
thanks!

The side lengths are $x, x, y$.

The volume is

$V = x^2y$

So $8 = x^2y$

$y = \frac{8}{x^2}$.

The surface area is

$TSA = 2x^2 + 4xy$

So $A = 2x^2 + 4xy$

$A = 2x^2 + 4x\left(\frac{8}{x^2}\right)$

$A = 2x^2 + \frac{32}{x}$.
• Dec 27th 2009, 02:24 AM
e^(i*pi)
Quote:

Originally Posted by BabyMilo
thanks!

What is known:

Code:

 <br /> img.top {vertical-align:15%;}<br /> $V = 8m^3$ square base. <br /> img.top {vertical-align:15%;}<br /> $A_s = Am^2$
Use the formula for the volume of a cube, given that you have a square base an $x^2$ term should appear. It will be easiest to introduce a new variable for the height - $h$ in my case.

Spoiler:
$V = x^2h = 8 \: \: \rightarrow \: \: h = \frac{8}{x^2}$ where $h$ is the height of the cuboid

You can now get an expression for the surface area given that it is equal to the sum of it's faces. If you draw a diagram you should be able to find the expression more easily.

Spoiler:
$A = 2x^2 + 4xh$

You have simultaneous equations to solve now. Eliminate h as it's not being asked for and get the expression for A