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A non-square rectangle has integer dimensions. The number  of square units in its area is four times the number of units in its perimeter. What is the smallest possible perimeter of the rectangle?

 Aug 8, 2023
 #1
avatar+42 
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I am editing this answer in order to make it more readable.

It does look computer generated but I personally don't really care, except that where an answer comes from should ALWAYS be cited.

The answer appears to be a good one.

Melody 

(all my changes are clearly marked)

 

Let’s assume the length of the rectangle is L and the width is W.
The area of the rectangle is given by A = L * W. The perimeter of the rectangle is given by P = 2L + 2W.
According to the problem, the number of square units in the area is four times the number of units in the perimeter. Mathe- matically, this can be represented as:
A = 4P
Substituting the expressions for A and P, we get:
L * W = 4(2L + 2W)

L * W = 8L + 8W

Rearranging the equation, we get:
L * W - 8L - 8W = 0
We can rewrite this equation as:
LW - 8L - 8W = 0
To make it easier to solve, let’s add 64 to both sides of the equation:
LW - 8L - 8W + 64 = 64

LW - 8L - 8W + 64 = 64
Now, we can factor the left side of the equation:
(L - 8)(W - 8) = 64


We are looking for the smallest possible perimeter, which means we want to minimize the sum of the length and width.

Therefore, we need to find the smallest possible values for L and W that satisfy the equation.

 

The factors of 64 are: 1, 2, 4, 8, 16, 32, 64.
We can try different combinations of     L - 8    and    W - 8    to see which ones give us integer values for L and W.

If L - 8 = 1 and W - 8 = 64, we get L = 9 and W = 72, which are not integer values.  (I think you mean that they are integer values)

If L - 8 = 2 and W - 8 = 32, we get L = 10 and W = 40, which are integer values.
If L - 8 = 4 and W - 8 = 16, we get L = 12 and W = 24, which are integer values.
If L - 8 = 8 and W - 8 = 8, we get L = 16 and W = 16, which are integer values.
 

So far what you have said appears to make sense  -  But it also looks like it has been computer generated.

Therefore, the smallest possible perimeter of the rectangle is 2L + 2W = 2(10) + 2(40) = 20 + 80 = 100 units. You lose me here - Melody

 

Added by Melody.

I think this gives possible perimeters of    

2(9+72)      = 162

2(10+40)     = 100

2(12+24)     = 72

2(16+16)     = 64  ( but this one is a square which is contrary to the conditions. )

 

So the smallest perimeter appears to be 72. 

This happens when the length is 24 and the width is 12

Area = 288 u^2

Perimeter = 72

72* 4 = 288      excellent     laugh   

 Aug 8, 2023
edited by Melody  Aug 9, 2023
 #3
avatar+118623 
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Some people were claiming that this was a nonsence answer so I decided to take a look.

It is actually a very good answer except for some poor formatting and a couple of 'silly' errors.

Next time Hairy, please cite your source.    laugh

 

Thanks for your answer too guest  laugh

Melody  Aug 9, 2023
 #4
avatar+42 
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It's actually not computer generated. I usually write words next each other. If it was computer generated than I would've cited that. I explain in terms that make it look like it's computer generated. 

IhaveHairystuff  Aug 9, 2023
 #2
avatar
+1

Area = L * W

 

Perimeter =2L  +  2W

 

L * W = 4 * [2L  +  2W]

L * W = 8L  +  8W

 

By trying a few numbers, it turns out that the smallest perimeter is:

When L = 24  and W = 12

 

Then Perimeter is: [2*L  +  2*W] =[2 *24  +  2 * 12] = 72

 

And the area =24  *  12 = 288 u^2

 

And: 4  * 72 = 288 = The area

 Aug 9, 2023

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