The perimeter of a rectangle and square are equal. If the length and breadth of the rectangle are 32 cm and 18 cm respectively, find which f

Question

The perimeter of a rectangle and square are equal. If the length and breadth of the rectangle are 32 cm and 18 cm respectively, find which figure has greater area and by how much?

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1 month 2021-10-26T12:24:24+00:00 2 Answers 0 views 0

If the perimeter of a square and a rectangle is equal, which of the two has a greater area?

Don’t let change and uncertainty hold you back.

There are definitely other ways to do this, many of which are simpler and more intuitive. But screw it, let’s prove this beyond a shadow of a doubt for all possibilities. In other words, let’s answer this with some basic calculus. That’s right, everybody, we’re going to do an optimization!! YAY!! (The crowd goes wild)

Say we’ve got some rectangular shape with dimensions x by y. The perimeter will therefore equal 2x + 2y. If we determine that the perimeter has some constant value of p, we can say that 2x + 2y = p. We can also therefore define y in terms of x, stating that y = p/2 – x

Now that we have defined y in terms of x, we can state that the area (A) of the rectangle will equal x (p/2 – x).

In order to do an optimization, you want to take the first derivative of a function and then find the critical points.

Our initial equation:

A = px/2 – x^2

first derivative

A’ = p/2 – 2x

Critical points occur when the first derivative equals zero. So let’s solve for x

0 = p/2 – 2x

x = p/4

So when x = p/4, we will either be at a maximum or minimum area. In order to determine which, we pick out points on either side. If p/4 is a minimum, than any values less than p/4 plugged into our derivative equation will be positive, showing that the area would be increasing there, and any values above p/4 would give us negative terms in our derivative, showing us that our area was decreasing.

Let’s choose some easy values, 0 and p/2.

f’(0) = p/2 – 2(0) → +

f’(p/2) = p/2 – 2(p/2) = p/2 – p = -p/2 → –

Our quick derivative test tells us that p/4 is a maximum for our area. So if x were to equal p/4, what would y equal?

Remember, y = p/2 – x

y = p/2 – p/4

y = 2p/4 – p/4

y = p/4

This tells us that when the area is at its maximum, both x and y will equal p/4. Therefore, a square will have a greater area than any non-rhombus rectangle with the same perimeter.

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Perimeter of the rectangle = 2(l + b) units

● Length of the rectangle = P2 – b units

● Breadth of the rectangle = P2 – l units

● Area of the rectangle = l × b sq. units.

● Length of the rectangle = Ab units .

● Breadth of the rectangle = Al units

● Diagonal of the rectangle = l2+b2−−−−−√ units

Let us consider a rectangle of length ‘a’ units and breadth ‘b’ units.

Perimeter of Rectangle

Therefore, perimeter of the rectangle ABCD

= (AB + BC + CD + DA) units

= (a + b + a + b) units

= (2a + 2b) units

= 2 (a + b) units

Therefore, perimeter of the rectangle = 2 (length + breadth) units

We know that the area of the rectangle is given by