SOLUTION: Please help me with my assignment for investigating the relationship between the width and the area of rectangles with a constant perimeter of 24cm

Algebra ->  Rectangles -> SOLUTION: Please help me with my assignment for investigating the relationship between the width and the area of rectangles with a constant perimeter of 24cm       Log On


   

Question 1136417: Please help me with my assignment for investigating the relationship between the width and the area of rectangles with a constant perimeter of 24cm
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
let l be the length of the rectangle and w be its width, then we have
:
2l +2w = 24
:
l + w = 12
:
solve for l
:
l = 12 -w
:
Area of rectangle(A) = l * w
:
substitute for l
:
A = (12 -w) * w
:
************************************************
Relationship between area and width of rectangle
:
A = 12w -w^2
************************************************
:
the graph of this equation is a parabola that opens downward, the x-axis is values for width and the y axis is the area of the rectangle
:
+graph%28+300%2C+200%2C+-2%2C+13%2C+-20%2C+40%2C+-x%5E2+%2B12x%29+
:
make a table
:
width Area ratio width to area
----- ---- -------------------
1.....11...1/11
2.....20...1/10
3.....27...1/9
4.....32...1/8
5.....35...1/7
6.....36...1/6
7.....35...1/5
8.....32...1/4
9.....27...1/3
10....20...1/2
11....11...1
:
Maximum area of the rectangle occurs when the width is 6cm and the length is 6cm
:
Note all squares are rectangles but all rectangles are not squares
: