A cattle rancher wants to enclose a rectangular area and then divide it into five pens with fencing parallel to one side of the rectangle (see the figure below). There are 560 feet of fencing available to complete the job. What is the largest possible total area of the five pens?
3 answers
So there are two long sides and 6 short sides. Maximum area will be when the fencing is divided equally between the two dimensions. In this case, that would be 280 ft each. So the area will be 140 by 46.67 = 6533.33 ft^2
2 L + 6 W = 560 ... L + 3 W = 280
L = 280 - 3 W
a = L * W = 280 W - 3 W^2
amax is on the axis of symmetry
Wmax = -280 / (2 * -3)
L = 280 - 3 W
a = L * W = 280 W - 3 W^2
amax is on the axis of symmetry
Wmax = -280 / (2 * -3)
Probable solution expected by the instructor:
let the length of the whole compound be y
let the width of each pen be x
given: 2y + 6x = 560
y = 280 - 3x
area = xy
= x(280-3x) = 280x - 3x^2
d(area)/dx = 280 - 6x
= 0 for a max of area
6x = 280
x = 280/6 = 140/3
y = 280 - 3(140/3) = 140
max area = (140)(140/3) = 19600/3
the max area is 19600/3 ft^2
let the length of the whole compound be y
let the width of each pen be x
given: 2y + 6x = 560
y = 280 - 3x
area = xy
= x(280-3x) = 280x - 3x^2
d(area)/dx = 280 - 6x
= 0 for a max of area
6x = 280
x = 280/6 = 140/3
y = 280 - 3(140/3) = 140
max area = (140)(140/3) = 19600/3
the max area is 19600/3 ft^2
2 answers