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...

square with two sides labeled 4m

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!):

rectangle with the long side labeled 6m and the short side labeled 2m

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...

square with two sides labeled 7ft.

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!).

rectangle with one side labeled 9ft and the short side labeled 5ft

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:

rectangle with both short sides labeled "x"

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.

rectangle with one side labeled "x" and the long side labeled "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!

parabola on a graph with its max point marked at (4, 16)

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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