Ed is planning to put up a rectangular garden with a fixed area of 120m^2. If the dimension of the garden have to be whole numbers,determine the dimension that will require the least amount of fencing materials to enclose the garden.

Length = X meters

Width = x-1 meters

x * (x-1) = 120.
x^2 - x = 120,
x^2 - x - 120 = 0,
Use Quad. Formula.
X = (-B +- sqrt(B^2-4AC))/2A. m.
X = (1 +- sqrt(1 + 480))/2 = 11.5 and -10.5.
X = 12 m. = Length.
L * W = 120,
12 * W = 120,
W = 10 m.

What measurements would make this closest to a square?

x y = 120

120 = 12*10
11 * 11 would not quite work, close though :)
would you use 6*20 ????

10 and 12?

I need algebraic solution for this hehe I don't know how

@damon, do youu know algebraic process of it?

Well, if Ed wants to minimize the amount of fencing material needed, he should probably go with a square garden. Why? Because a square garden will have all sides equal, which means less fencing required... and less time spent arguing over which side is longer. So, Ed should go for a garden that is 10 meters by 12 meters, since 10 times 12 equals 120. It's simple, symmetrical, and perfectly proportional... just like a clown's red nose.

To determine the dimensions that will require the least amount of fencing materials to enclose the garden, we need to consider the possible dimensions that result in a fixed area of 120 square meters.

Let's start by finding the possible length and width values that have a product of 120.

As the dimensions have to be whole numbers, we can list all the possible pairs of factors of 120:

1 × 120 = 120
2 × 60 = 120
3 × 40 = 120
4 × 30 = 120
5 × 24 = 120
6 × 20 = 120
8 × 15 = 120
10 × 12 = 120

Now, let's calculate the perimeter (fencing materials required) for each pair of dimensions:

Perimeter of rectangle = 2(length + width)

For the first pair, (1, 120):
Perimeter = 2(1 + 120) = 2(121) = 242

Similarly, we can calculate the perimeter for each pair:

(2, 60): Perimeter = 2(2 + 60) = 2(62) = 124
(3, 40): Perimeter = 2(3 + 40) = 2(43) = 86
(4, 30): Perimeter = 2(4 + 30) = 2(34) = 68
(5, 24): Perimeter = 2(5 + 24) = 2(29) = 58
(6, 20): Perimeter = 2(6 + 20) = 2(26) = 52
(8, 15): Perimeter = 2(8 + 15) = 2(23) = 46
(10, 12): Perimeter = 2(10 + 12) = 2(22) = 44

From the calculations, the dimensions with the least amount of fencing materials required are (10, 12), resulting in a perimeter of 44. Therefore, the garden with dimensions 10m × 12m will require the least amount of fencing materials to enclose.