Finding the length and width of rectangle

That's essentially what you're doing here. But it's not a good approach when you're first learning how to write proofs. It's better to stick with "direct proof" (starting with what you know and working towards the conclusion) until you're more experienced.
  • #1
chwala
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Mod note: Thread originally posted in a technical math section, so is missing the homework template.
I am trying to solve this problem it states as follows "A piece of wire 24 cm long has the shape of a rectangle.
(a) Given that the width is W cm, show that the area, A cm^2 of the rectangle is given by the function A=36-(6-W)^2. (b) Find the greatest possible domain and corresponding range of the function.

My challenge is in part (a) this is how i attempted it
2L+2W= 24 , L+W=12 , L=12-W , ...W(12-W)= 36-(6-W)^2...where 12W-W^2=36-36+12W-W^2, whence 12W-W^2=12W-W^2 this is correct but is this the way to show it?:L
 
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  • #2
chwala said:
whence 12W-W^2=12W-W^2

This part is obviously unhelpful.

You have that

[tex]12w-w^2=36-36+12w-w^2[/tex]

Now you should do

[tex]=36-(w^2-12w+36)[/tex]
[tex]=36-(w-6)^2[/tex]
[tex]=36-(6-w)^2[/tex]

or if you're proficient enough at spotting binomials (you already know where you are going to end up so you don't even really need to be in this case), then just simply do

[tex]12w-w^2=36-36+12w-w^2[/tex]
[tex]=36-(36-12w+w^2)[/tex]
[tex]=36-(6-w)^2[/tex]
 
  • #3
I appreciate but you have just re- expressed the rhs which we have already been given, in the question... we are supposed to come up with the left hand side expression which should be equal to A=36-(6-W)^2 (the rhs)
 
  • #4
What are you referring to when you say "left hand expression"?
 
  • #5
##12w-w^2##
 
  • #6
chwala said:
I appreciate but you have just re- expressed the rhs which we have already been given, in the question... we are supposed to come up with the left hand side expression which should be equal to A=36-(6-W)^2 (the rhs)

You did come up with the correct equation for the area in terms of the width, which is A=w(12-w). Now you just need to manipulate this expression to show that it is in fact equivalent to 36-(6-w)2. Once you've done that, you've answered the question.
 
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  • #7
chwala said:
(a) Given that the width is W cm, show that the area, A cm^2 of the rectangle is given by the function A=36-(6-W)^2. (b) Find the greatest possible domain and corresponding range of the function.

My challenge is in part (a) this is how i attempted it
2L+2W= 24 , L+W=12 , L=12-W ,
This is absolutely correct. Then you are asked to show: A= W*L = W(12-W) is equivalent to 36-(6-W)^2.
I think you did this too. You expanded the right hand side and showed it was equal to the left hand side.

...W(12-W)= 36-(6-W)^2...where 12W-W^2=36-36+12W-W^2, whence 12W-W^2=12W-W^2 this is correct but is this the way to show it?:L

If you wanted to start from the left hand side and manipulate it to show that you can get the right hand side, just add zero in the form of 36-36. Then regroup.
Either way, as long as you present it clearly, you have shown that the two expressions are equivalent.
 
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  • #8
chwala said:
Mod note: Thread originally posted in a technical math section, so is missing the homework template.
I am trying to solve this problem it states as follows "A piece of wire 24 cm long has the shape of a rectangle.
(a) Given that the width is W cm, show that the area, A cm^2 of the rectangle is given by the function A=36-(6-W)^2. (b) Find the greatest possible domain and corresponding range of the function.

My challenge is in part (a) this is how i attempted it
2L+2W= 24 , L+W=12 , L=12-W , ...W(12-W)= 36-(6-W)^2...where 12W-W^2=36-36+12W-W^2, whence 12W-W^2=12W-W^2 this is correct but is this the way to show it?:L
There are better ways than what you wrote. First, don't start with W(12-W) = 36-(6-W)2 because that's what you're supposed to prove. If you start off by writing that down, you're assuming what's to be shown, i.e., begging the question. Second, what you wrote means:
\begin{align}
W(12-W) &= 36-(6-W)^2 \Rightarrow \\
12W-W^2 &= 36-36+12W-W^2 \Rightarrow \\
12W-W^2 &= 12W-W^2.
\end{align} It's not correct logically to start with some statement, reach a true conclusion, and infer that the initial statement was true. You could fix it in this case by simply pointing out each step is reversible, but I think that's kind of clumsy. In problems like this, it's better to start with one side and manipulate it into being equal to the second side, which is what Mentallic did above.
 
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  • #9
Starting with what you want to prove and using invertible steps to arrive at an "obviously true" conclusion (say, one of your hypotheses) is called "synthetic proof" and is often used.
 
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Related to Finding the length and width of rectangle

1. How do you find the length and width of a rectangle?

The length and width of a rectangle can be found by measuring two adjacent sides of the rectangle using a ruler or measuring tape. The length is the longer side, while the width is the shorter side.

2. What is the formula for finding the area of a rectangle?

The formula for finding the area of a rectangle is length x width. This means you multiply the length of the rectangle by its width to get the total area.

3. Can you use the Pythagorean theorem to find the length and width of a rectangle?

No, the Pythagorean theorem is used to find the length of the hypotenuse in a right triangle. It cannot be directly applied to finding the length and width of a rectangle.

4. How do you find the dimensions of a rectangle if you only know the area?

If you know the area of a rectangle, you can find the dimensions by dividing the area by either the length or width. For example, if the area is 20 square units and the length is 5 units, the width would be 4 units (20/5 = 4).

5. What are some real-life applications of finding the length and width of a rectangle?

Finding the length and width of a rectangle is useful in many real-life situations, such as measuring the dimensions of a room or furniture, calculating the size of a rug or carpet, or determining the size of a garden or plot of land.

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