# Wire of Length x

• Nov 3rd 2008, 07:08 AM
magentarita
Wire of Length x
A wire of length x is bent into the shape of a square.

(a) Express the perimeter of the square as a function of x.

(b) Express the area of the square as a function of x.
• Nov 3rd 2008, 07:24 AM
Peritus
1. P = x
2. S = (x/4)^2
• Nov 4th 2008, 06:45 AM
magentarita
ok but...
Quote:

Originally Posted by Peritus
1. P = x
2. S = (x/4)^2

• Nov 4th 2008, 08:12 AM
masters
Quote:

Originally Posted by magentarita

A wire of length x bent into a square would make each side $\frac{1}{4}x$

Perimeter of the square $P(x)=\frac{1}{4}x+\frac{1}{4}x+\frac{1}{4}x+\frac{ 1}{4}x=\frac{4}{4}x=x$

Area of the square would be $A(x)=\left(\frac{1}{4}x\right)^2=\frac{1}{16}x^2$
• Nov 4th 2008, 09:50 PM
magentarita
wonderful...........
Quote:

Originally Posted by masters
A wire of length x bent into a square would make each side $\frac{1}{4}x$

Perimeter of the square $P(x)=\frac{1}{4}x+\frac{1}{4}x+\frac{1}{4}x+\frac{ 1}{4}x=\frac{4}{4}x=x$

Area of the square would be $A(x)=\left(\frac{1}{4}x\right)^2=\frac{1}{16}x^2$