Asked by Razel
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.
Answers
Answered by
Ms. Sue
What measurements would make this closest to a square?
Answered by
Razel
10 and 12?
I need algebraic solution for this hehe I don't know how
I need algebraic solution for this hehe I don't know how
Answered by
Damon
x y = 120
120 = 12*10
11 * 11 would not quite work, close though :)
would you use 6*20 ????
120 = 12*10
11 * 11 would not quite work, close though :)
would you use 6*20 ????
Answered by
Daryl
@damon, do youu know algebraic process of it?
Answered by
Henry2
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.
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.
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