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