Find Perimeter of Rectangle with Given Area and Width

[mathdad]

Given the width of a rectangle whose area is 25ft^2 is x, find the perimeter of this rectangle.

Let me see.

A = L•W

25 = L•x

25/x = L

P = 2L + 2W

P = 2(25/x) + 2x

P = (50/x) + 2x

Is this right?

If it is right, use the inequality below to show that P ≥ 20.

In the following inequality, a and b are positive numbers.

sqrt{ab} ≤ (a + b)/2

In words, the above inequality means THE GEOMETRIC MEAN IS LESS THAN OR EQUAL TO THE ARITHMETIC MEAN.

 

[MarkFL]

Given the width of a rectangle whose area is 25ft^2 is x, find the perimeter of this rectangle.

Let me see.

A = L•W

25 = L•x

25/x = L

P = 2L + 2W

P = 2(25/x) + 2x

P = (50/x) + 2x

Is this right?

Yes, this is correct.

If it is right, use the inequality below to show that P ≥ 20.

In the following inequality, a and b are positive numbers.

sqrt{ab} ≤ (a + b)/2

In words, the above inequality means THE GEOMETRIC MEAN IS LESS THAN OR EQUAL TO THE ARITHMETIC MEAN.

Okay, we would have:

\(\displaystyle \sqrt{LW}\le\frac{L+W}{2}\)

Can you continue?

 

[mathdad]

Do I plug (25/x) for L and x for W?

The problem then becomes another inequality question, right?

 

[MarkFL]

Do I plug (25/x) for L and x for W?

The problem then becomes another inequality question, right?

No, you plug in 25 for LW and P/4 for (L + W)/2...can you work out the implications for a general rectangle?

 

[mathdad]

sqrt{25} = P/4

5 = P/4

P = 4(5)

P = 20

Where did P/4 come from?

I know that LW comes from (25/x)(x/1) = 25.

 

[MarkFL]

sqrt{25} = P/4

5 = P/4

P = 4(5)

P = 20

Where did P/4 come from?

I know that LW comes from (25/x)(x/1) = 25.

Making the suggested substitutions, you actually get:

\(\displaystyle \sqrt{25}\le\frac{P}{4}\implies 20\le P\)

This is what you were asked to show. The fact that:

\(\displaystyle LW=25\)

simply comes from the fact that length times width is area and the area was given as 25. Then, observe that we have:

\(\displaystyle P=2(L+W)\implies \frac{P}{4}=\frac{L+W}{2}\)

 

[mathdad]

I will work on this on my next day off.

 

[MarkFL]

Let's examine a general rectangle...using the AM - GM inequality we may state:

\(\displaystyle \sqrt{LW}\le\frac{L+W}{2}\)

When does equality occur?

Given that the perimeter P is:

\(\displaystyle P=2(L+W)\)

We may write:

\(\displaystyle 4\sqrt{LW}\le P\)

Using the condition of equality we found above, what can we infer from this?

 

[mathdad]

I am off tomorrow and will get back to this question and both quadratic inequality questions on Friday.

 

[mathdad]

Let's examine a general rectangle...using the AM - GM inequality we may state:

\(\displaystyle \sqrt{LW}\le\frac{L+W}{2}\)

When does equality occur?

Given that the perimeter P is:

\(\displaystyle P=2(L+W)\)

We may write:

\(\displaystyle 4\sqrt{LW}\le P\)

Using the condition of equality we found above, what can we infer from this?

Sorry but I don't follow what you're saying. Can you complete the prove step by step?

 

[MarkFL]

Sorry but I don't follow what you're saying. Can you complete the prove step by step?

We have:

\(\displaystyle \sqrt{LW}\le\frac{L+W}{2}\)

When I ask "when does equality occur," I am essentially asking you to solve:

\(\displaystyle \sqrt{LW}=\frac{L+W}{2}\)

Now the two sides are set equal to each other...what do you get when you solve this?

 

[mathdad]

sqrt{LW} = (L + W)/2

[sqrt{LW}]^2 = [(L + W)/2]^2

LW = (L + W)^2/4

QUESTION:

AM I SOLVING FOR L OR W?

 

[MarkFL]

You're just seeing what the equation implies about the relationship between L and W. :D

 

[mathdad]

I cannot see the connection between L, W, and the given inequality.

Must I simplify a little more?

 

[MarkFL]

I cannot see the connection between L, W, and the given inequality.

Must I simplify a little more?

Yes, once you simplify, you will find the implication. :D

 

[mathdad]

LW = (L + W)^2/4

4LW = (L + W)^2

4LW = (L + W)(L + W)

4LW = L^2 + 2LW + W^2

L^2 + 2LW - 4LW + W^2 = 0

L^2 - 2LW + W^2 = 0

(L + W)^2 = 0

What is next?

 

[MarkFL]

LW = (L + W)^2/4

4LW = (L + W)^2

4LW = (L + W)(L + W)

4LW = L^2 + 2LW + W^2

L^2 + 2LW - 4LW + W^2 = 0

L^2 - 2LW + W^2 = 0

Your next step should be:

\(\displaystyle (L-W)^2=0\)

So, how must L and W be related in order for this to be true?

 

[mathdad]

They are related quadratically.

 

[MarkFL]

They are related quadratically.

What we see is that we must have:

\(\displaystyle L=W\)

And so, what we have found is that for a rectangle of a given area, the perimeter is the smallest it can be when L = W...in other words, for a given rectangle of area A, the perimeter is minimized when the rectangle is a square. This hints at the power of the AM - GM inequality, which can sometimes be used to solve a problem that would ordinarily require calculus. :D

 

[mathdad]

You are amazing.

I could solve (L - W)^2 = 0, find L to equal W and it would mean nothing to me.

Are you familiar with David Cohen's Precalculus With Unit Circle Trigonometry 3rd Edition?

It is simply one of the most challenging precalculus textbooks out there.

In terms of this question, I had no idea that it is related to calculus.

 

[MarkFL]

You are amazing.

I could solve (L - W)^2 = 0, find L to equal W and it would mean nothing to me.

When I was a student, I found one of the best ways to understand things is to look at them in general terms. For example, in this problem, rather than looking at a rectangle having a set area, we can look at all rectangles by using A for the area rather than 25.

Are you familiar with David Cohen's Precalculus With Unit Circle Trigonometry 3rd Edition?

It is simply one of the most challenging precalculus textbooks out there.

I have Cohen's 3rd edition Pre-Calc textbook, although the title of mine is:

Precalculus: A Problems-Oriented Approach - 3rd Edition

In terms of this question, I had no idea that it is related to calculus.

Yes, optimization (minimization/maximization) is something studied in differential calculus. ;)

 

[mathdad]

Good to know that you have the same book. All my questions should be in your textbook, right? From now on, I will post the section, page & question numbers. There are so many great math questions in David's book. Wish I had time to post every EVEN NUMBER question. Even number question answers are not given in his book and in most math textbooks. Even number questions are done in most classrooms.