A piece of wire 68cm long in length is bent into the shape of a rectangle.

a.) If x cm is the length of the rectangle and A cm^2 is the area enclosed by the rectangular shape, write down a formula which connects A and x.

b.) Sketch the graph of A against x for suitable x-values.

c.) Use your graph to determine the maximum area formed.

2. Originally Posted by andrew2322

A piece of wire 68cm long in length is bent into the shape of a rectangle.

a.) If x cm is the length of the rectangle and A cm^2 is the area enclosed by the rectangular shape, write down a formula which connects A and x.

b.) Sketch the graph of A against x for suitable x-values.

c.) Use your graph to determine the maximum area formed.
You choose x as the side of a rectangle with perimeter 68. Than your other side will have length (68-2x)/2 = 34-x. The reason for this is that, if x is one side of a rectangle, than the length of that side plus the length of its opposite side will be 2x. Then the remaining perimeter is 68-2x. This gives the total length of the two opposite sides remaining. Half of this length is the length of the other side. This gives the (68-2x)/2 = 34-x.
The area of this rectangle is then A=x(34-x)=-x^2+34x. This is because x is one side, and the other is 34-x.
The only values of x that are possible are those between 0 and 34. If you graph the function -x^2+34x from 0 to 34, you will get the picture attached.

The graph will show that the Area is maximized at the point where x=17. The area corresponding to this point is 289. It should be noted that of all rectangles of equal perimeter, the one enclosing the maximum area will be exactly the rectangle with all sides equal, or in other words will be the square of that perimeter.

3. ## sorry

sorry im still really confused, i get the equation forming and all, but could you please expand on the graphing bit and finding the right answers? its all a big blur to me