Asked by aliff
David has 160 metres of fencing to build a rectangular corral. He wants to give his animals as much space as possible by using the length of fencing that he can afford. Find the dimension and the maximum area of the corral.
Answers
Answered by
mathhelper
Common sense will tell you that the enclosure would have to be a square,
so if each side is x, then 4x = 160
x = 40 m
area = 40^2 m^2 = 1600 m^2
but ....
let the width be x and the length be y m
then 2x+2y = 160
x+y = 80, or y = 80-x
area = xy = x(80-x) - -x^2 + 80x
this is a quadratic and would graph as a parabola
the x of the vertex is -b/(2a) = -80/-2 = 40
if x = 40, then y = 80-40 = 40
area = xy = 40*40 = 1600 m^2
(as anticipated above)
so if each side is x, then 4x = 160
x = 40 m
area = 40^2 m^2 = 1600 m^2
but ....
let the width be x and the length be y m
then 2x+2y = 160
x+y = 80, or y = 80-x
area = xy = x(80-x) - -x^2 + 80x
this is a quadratic and would graph as a parabola
the x of the vertex is -b/(2a) = -80/-2 = 40
if x = 40, then y = 80-40 = 40
area = xy = 40*40 = 1600 m^2
(as anticipated above)
Answered by
R_scott
the largest area to perimeter ratio for a rectangle is a square
L + W = 80 ... L = 80 - W
A = L * W = 80 W - W^2
the max is on the axis of symmetry of the parabola
... Wmax = - b / (2 a) = -80 / (2 * -1) = 40
Lmax = 80 - Wmax = 40
Amax = Lmax * Wmax = 40 * 40 = 1600 m^2
L + W = 80 ... L = 80 - W
A = L * W = 80 W - W^2
the max is on the axis of symmetry of the parabola
... Wmax = - b / (2 a) = -80 / (2 * -1) = 40
Lmax = 80 - Wmax = 40
Amax = Lmax * Wmax = 40 * 40 = 1600 m^2
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