contestada

A rancher wants to fence in a rectangular area of 24600 square feet in a field and then divide the region in half with a fence down the middle parallel to one side. What is the smallest length of fencing that will be required to do this?

Respuesta :

elliottou

Answer:

120[tex]\sqrt{41}[/tex] ft of fencing

Step-by-step explanation:

24600 = xy

Fence (F) = 3x + 2y

24600/y  = x

F = 3(24600/y) + 2y

F = 73800/y + 2y

dF/dy = -73800/y^2 + 2

2 = -73800/y^2

y^2 = -73800/2 = 36900

y = 30[tex]\sqrt{41}[/tex]

24600 = x(30[tex]\sqrt{41}[/tex])

24600 / 30[tex]\sqrt{41}[/tex] = x = 20[tex]\sqrt{41}[/tex]

total fencing = 3(20[tex]\sqrt{41}[/tex]) + 2(30[tex]\sqrt{41}[/tex]) = [tex]\sqrt{41}[/tex](60+60) = 120[tex]\sqrt{41}[/tex]

hafsaabdulhai

Answer:

784.2 ft

Step-by-step explanation:

Area= xy

Smallest paramenter and so smallest fencing will occur when x=y

Area= x²

24,600= x²

x= 156.84

Length of fencing= 5x

                       = 5(156.84)

                       = 784.2 ft