Maximizing the Area of a Rectangle
The problem
A farmer has 100 metres of fencing and would like to make a rectangular enclosure in which to keep his horses.
He wants the enclosure to have the largest possible area and would like to know what size sides the enclosure should have to make this possible.
He wants the enclosure to have the largest possible area and would like to know what size sides the enclosure should have to make this possible.
The area of a rectangle
For any rectangle, the area is calculated by multiplying the length by the width e.g. a rectangle of 10 meters by 20 meters would have an area of 10 x 20 = 200 m^2.
The perimeter is found by adding all of the sides together (i.e. how much fence is needed to go around the rectangle). For the rectangle mentioned above, the perimeter = 10 + 20 + 10 + 20 = 60 m.
The perimeter is found by adding all of the sides together (i.e. how much fence is needed to go around the rectangle). For the rectangle mentioned above, the perimeter = 10 + 20 + 10 + 20 = 60 m.
Which rectangle is best?
The farmer starts off by creating an enclosure measuring 30 metres by 20 metres. He has used all of the fencing as 30 + 20 + 30 + 20 = 100m and he has got an area of 30 x 20 = 600m^2.
He then decides that he can probably create a larger area if he makes the rectangle longer. He makes an enclosure that is 40 meters long. Unfortunately, as the enclosure is now longer, he is running out of fencing (and so it is now only 10 meters wide). The new area is 40 x 10 = 400m^2. The longer enclosure is smaller than the first one.
Wondering if there is a pattern to this, the farmer makes an even longer, thinner enclosure of 45 meters by 5 meters. This enclosure has an area of 45 x 5 = 225m^2, even smaller than the last one. There definitely seems to be a pattern here.
To try to create a larger area, the farmer then decides to go the other way and make the enclosure shorter again. This time he takes it to the extreme of the length and width being the same size: a square of 25 metres by 25 metres.
The square enclosure has an area of 25 x 25 = 625 m^2. This is definitely the biggest area so far, but being a thorough person, the farmer would like to prove that he has found the best solution. How can he do this?
He then decides that he can probably create a larger area if he makes the rectangle longer. He makes an enclosure that is 40 meters long. Unfortunately, as the enclosure is now longer, he is running out of fencing (and so it is now only 10 meters wide). The new area is 40 x 10 = 400m^2. The longer enclosure is smaller than the first one.
Wondering if there is a pattern to this, the farmer makes an even longer, thinner enclosure of 45 meters by 5 meters. This enclosure has an area of 45 x 5 = 225m^2, even smaller than the last one. There definitely seems to be a pattern here.
To try to create a larger area, the farmer then decides to go the other way and make the enclosure shorter again. This time he takes it to the extreme of the length and width being the same size: a square of 25 metres by 25 metres.
The square enclosure has an area of 25 x 25 = 625 m^2. This is definitely the biggest area so far, but being a thorough person, the farmer would like to prove that he has found the best solution. How can he do this?
Proof that the square is the best solution
To prove that the square is the best solution, the farmer decides to use some algebra. He denotes one side with the letter x. He then works out an expression for the other side in terms of x. The perimeter is 100m and we have two opposite sides that have length x, so 100 - 2x gives us the total of the other two sides. As these two sides are the same as each other, halving this expression will give us the length of one of them so (100 - 2x) ÷ 2 = 50 - x. We now have a rectangle of width x and length 50 - x.
A rectangle with algebraic side lengths
Finding the optimal solution
The area of our rectangle is still length × width so:
Area = (50 - x) × x
= 50x - x^2
To find maximum and minimum solutions to an algebraic expression we can use differentiation. By differentiating the expression for the area with respect to x, we get:
dA/dx = 50 - 2x
This is at a maximum or minimum when dA/dx = 0 so:
50 - 2x = 0
2x = 50
x = 25m
Therefore our square is either a maximum solution or a minimum solution. As we already know that it is bigger than other rectangle areas that we have calculated, we know it cannot be a minimum, hence the biggest rectangular enclosure the farmer can make is a square of sides 25 meters with an area of 625m^2.
Area = (50 - x) × x
= 50x - x^2
To find maximum and minimum solutions to an algebraic expression we can use differentiation. By differentiating the expression for the area with respect to x, we get:
dA/dx = 50 - 2x
This is at a maximum or minimum when dA/dx = 0 so:
50 - 2x = 0
2x = 50
x = 25m
Therefore our square is either a maximum solution or a minimum solution. As we already know that it is bigger than other rectangle areas that we have calculated, we know it cannot be a minimum, hence the biggest rectangular enclosure the farmer can make is a square of sides 25 meters with an area of 625m^2.
Are we sure the square is best?
But is a square the best solution of all? So far, we have only tried rectangular enclosures. What about other shapes?
If the farmer made his enclosure into a regular pentagon (a five-sided shape with all sides the same length) then the area would be 688.19 m^2. This is actually bigger than the area of the square enclosure.
What about if we try regular polygons with more sides?
Regular hexagon area = 721.69 m^2.
Regular heptagon area = 741.61 m^2.
Regular octagon area = 754.44 m^2.
There is definitely a pattern here. As the number of sides increases, the area of the enclosure also increases.
Each time we add a side to our polygon, we get closer and closer to having a circular enclosure. Let's work out what the area of a circular enclosure with a perimeter of 100 metres would be.
If the farmer made his enclosure into a regular pentagon (a five-sided shape with all sides the same length) then the area would be 688.19 m^2. This is actually bigger than the area of the square enclosure.
What about if we try regular polygons with more sides?
Regular hexagon area = 721.69 m^2.
Regular heptagon area = 741.61 m^2.
Regular octagon area = 754.44 m^2.
There is definitely a pattern here. As the number of sides increases, the area of the enclosure also increases.
Each time we add a side to our polygon, we get closer and closer to having a circular enclosure. Let's work out what the area of a circular enclosure with a perimeter of 100 metres would be.
Area of a circular enclosure
We have a circle of perimeter 100 meters.
Perimeter = 2πr where r is the radius, so:
2πr = 100
πr = 50
r = 50/π
The area of a circle = πr^2, so using our radius we get:
Area = πr^2
= π(50/π)^2
= 795.55 m^2
which is considerably bigger than the square enclosure with the same perimeter!
Perimeter = 2πr where r is the radius, so:
2πr = 100
πr = 50
r = 50/π
The area of a circle = πr^2, so using our radius we get:
Area = πr^2
= π(50/π)^2
= 795.55 m^2
which is considerably bigger than the square enclosure with the same perimeter!
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