Question 1123737
A piece of wire 6 metres long is cut into two parts, one of which is used to form a square and the other to form a rectangle whose length is three times its width.
:
 If x is the length of the wire used to form the square:
then 
(6-x) = the length to form the rectangle
: 
A) find an expression for the area of the square.
A(x) = {{{(x/4)^2}}}
B) an expression for the area of the rectangle
let w = the width of the rectangle
then
3w = the width
and 
{{{3w^2}}} = the area
the relationship of w to (6-x); the perimeter of the rectangle
2w + 6w = (6-x)
8w = (6-x)
w = {{{(6-x)/8}}}
find the area in terms of x, (A = {{{3w^2}}}) replace with x
A(x) = {{{3((6-x)/8)^2}}} is the area of the rectangle
:
C) the lengths of the two parts of the sum of the areas is a minimum
A(x) = {{{(x/4)^2}}} + {{{3((6-x)/8)^2}}} 
the easiest way to find the minimum area is to graph it, y = the area
{{{ graph( 300, 200, -6, 8, -2, 4, (x/4)^2+(3((6-x)/8)^2))) }}}
minimum area when x = 2.5 m the length to form the square
and
6 - 2.5 = 3.5 m the length to form the rectangle