Place four pebbles on the sand in the form of a square. Keep adding
as few pebbles as necessary to double the area. How many extra
pebbles are added each time?
Investigate how this pattern of squares continues. You could
measure lengths, areas and angles.
Can you work out the area of the inner square and give an
explanation of how you did it?
This reasoning sounds right, well done. Don't worry that with a base and height of 10m the area fenced off makes a square; a square is just a special kind of rectangle! If we do a bit more thinking can we be absolutely certain there isn't some way of improving on 100m2? Nathan used the letters x and y to represent the base and the height of the
If the sides are length x and y metres, then the perimeter of the rectangle is 2x+2y metres. Since this is all the fence we can use we must have 2x+2y=40. The area of the rectangle is x times y which I can write as xy.
From the first equation we know that 2y=40-2x so that y=20-x.
Then the formula for the area can be written
xy = x(20-x) = 20x-x2.
The largest value the expression 20x-x2 can take is 100, which is when x=10.
This is exactly right. If you draw a graph of the function 20x-x2 you'll see that it is highest at x=10, with a value of 200. There are also advanced techniques which can help find the highest point if you can't draw a graph.
A really special symmetry argument goes like this: in the equations 2x+2y=40 and Area=xy, there is nothing different between x and y; you could swap their names and the equations would be the same. This often means that the values of x and y should be the same.
Braiden noticed the following, and made a list of possible solutions.
The second question gives you the same amount of fence sections but only three sides of a rectangle need to be filled.
The formula for the amount of fence used in the perimeter is x+2y=40.
The area is xy like before.
Notice x = 40-2y so we get:
Area = xy = (40-2y)y = 40y-2y2.
Now draw a graph or use advanced techniques to find the highest point of this function: 40y-y2. At y=10 we get an area of 200m2. Notice that in this case x=20, so the fenced area isn't a square.