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