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base (m) | height (m) | area (m2) |
10 | 10 | 100 |
9 | 11 | 99 |
5 | 15 | 75 |
1 | 19 | 19 |
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
rectangle.
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.