Question

A farmer is planning to build a rectangular vegetable garden. The length of the garden is always 3 meters more than twice the width. The farmer has enough fencing material to enclose a total perimeter of 60 meters.%0D%0A%0D%0AIn the context of this problem what are the constraints for the viable solutions?%0D%0A%0D%0A(1 point)%0D%0AResponses%0D%0A%0D%0ANo constraints%0D%0ANo constraints%0D%0A%0D%0AAll positive integers%0D%0AAll positive integers%0D%0A%0D%0AAll positive integers greater than zero%0D%0AAll positive integers greater than zero%0D%0A%0D%0AAll real numbers

Answer 1

To determine the constraints for the viable solutions in the context of the farmer's garden, we need to consider the relationships established in the problem.

1. The length \( L \) of the garden is defined as \( L = 2W + 3 \), where \( W \) is the width.
2. The total perimeter of the garden is given to be 60 meters. We know that the perimeter \( P \) of a rectangle is calculated as \( P = 2L + 2W \).

Substituting the expression for \( L \) into the perimeter equation:

\[
P = 2(2W + 3) + 2W = 4W + 6 + 2W = 6W + 6
\]

Setting this equal to the total perimeter:

\[
6W + 6 = 60
\]

Solving for \( W \):

\[
6W = 54 \\
W = 9
\]

Now using this width to find the length:

\[
L = 2(9) + 3 = 18 + 3 = 21
\]

Thus, the dimensions of the garden are 21 meters for length and 9 meters for width.

Now, considering the constraints:

- Both \( L \) and \( W \) must be positive, i.e., \( W > 0 \) and \( L > 0 \).
- Width \( W \) must also be a real number that meets the semicircular condition established by the fence's perimeter limit.

Thus, the constraints for viable solutions imply that both the width and length must be positive numbers.

Given the options provided, the appropriate response is:

**All positive integers greater than zero.**