MG - Solving Problems Using Geometric Methods Lesson
Solving Problems Using Geometric Methods
We can employ geometric methods as tools to solve a variety of design problems. One popular type of design problem that can easily be solved using geometric methods is maximizing area given a set of constraints. Let's investigate...
The square above has a perimeter of 4 m + 4 m + 4 m + 4 m = 4(4 m) = 16 meters.
On your own paper, sketch a rectangle that has the same perimeter (16 meters), but is NOT a square!
Here's mine (remember, yours can be different as long as it has a perimeter of 16 meters!):
What is the area of your rectangle? Solve it on your own paper.
Remember, for a rectangle, Area = (length)(width).
The area of my rectangle is A = (2)(6) = 12 meters2
Now let's look back at the original square. The area of a square is A = (side)(side).
A = (4 m)(4 m) = 16 meters2
What do you notice about this area compared to the area of my rectangle? What about the area of yours? It's larger than both! Coincidence? Let's try one more...
The square above has a perimeter of 7 ft. + 7 ft. + 7 ft. + 7 ft. = 4(7 ft.) = 28 feet.
On your own paper, sketch a rectangle that has the same perimeter (28 feet) but is NOT a square!
Here's mine (remember, yours can be different as long as it has a perimeter of 28 feet!).
What is the area of your rectangle? Solve it on your own paper.
Remember, for a rectangle, Area = (length)(width).
The area of my rectangle is A = (9)(5) = 45 feet2
Now let's look back at the original square. The area of a square is A = (side)(side).
A = (7 ft.)(7 ft.) = 49 feet2
What do you notice about this area compared to the area of my rectangle? What about the area of yours?
The area of each of our rectangles is smaller than the area of the square with the same perimeter. This is not a coincidence.
It is, in fact, true that the way to maximize a rectangle's area given a certain perimeter is to create a square!
What does this look like algebraically?
Let's use our first example with the required perimeter of 16 meters.
Let's say that x is the side length of one set of opposite sides. Then our rectangle looks like this:
The perimeter of a rectangle is 2(length) + 2(width).
For our rectangle, it looks like:
2(length) + 2(x) = 16
2(length) = 16 - 2x
length = 8 - x
So the length of our rectangle is 8 - x.
Now, if we want to maximize the area of this rectangle, we need to find the equation representing the area.
Area = (length)(width)
Area = (8 - x)(x)
Area = 8x - x2
Thus, our output - our y-coordinate - will represent area:
y = 8x - x2
We can graph this function and find the maximum... this will allow us to find the dimensions that maximize area!
Notice that the maximum of this parabola is (4, 16). This means that x = 4 and y = 16.
So...
Width: x = 4
Length: 8 - x = 8 - 4 = 4
Maximum area: y = 16
It will ALWAYS be the dimensions that create a square that maximize the area of a rectangle!
Check out the presentation below for some more ways we can use geometric methods to solve design problems! Be sure to take notes as you follow along!
Once you feel like you understand how to apply geometric methods to design problems, try the problems below to test yourself!
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