Question
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
Ms. Sue
What measurements would make this closest to a square?
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
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 ????
@damon, do youu know algebraic process of it?
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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