calculus question?

Question

A rancher wants to fence in an area of 3000000 square feet in a rectangular field and then divide it in half with a fence down the middle parallel to one side. What is the shortest length of fence that the rancher can use?
i really need help on this problem can someone help me?

Answer 1

Curious problem. Let x and y be the sides. Then

Area = xy = 3m, where m is million
Fence length = 3x + 2y

From the 1st equation, we know that

y = 3m / x
Fence length = 3x + 6m / x

Differentiate with respect to x, we get:

3 - 6m / x^2

Setting this to 0 for min/max, we get

0 = 3 - 6m / x^2
0 = x^2 - 2m
x - Sqrt (2m)
x = 1414 approximately
y = 2122 approximately


Addendum: bob shark is wrong

Answer 2

Calculus???---no

Take the Square root of 3000000 to get the length of the width and length of the field.

for example if the square feet was 4 you would have a length of 2 ft, width of 2 feet, a perimeter of 8 ft and if you wanted to add up all the fenceing to meet the perameters of the question , it (in my example is
Length (2 Ft x 2) = 4 ft + Width is (2 Ft x 2) = 4 Ft + distance to cut in half is 2 Feet...Total 10 ft of fencing.

So you get your Square root of 3000000 to get length of length and Width to make a square of equal sides and the fence to devide it will be the same as the length, or the width.

Answer 3

Ok, we have xy=3m (m for million sq.ft.) and the total length as L=perimeter+x=2(x+y)+x=3x+2y so L=3x+2y.

Combining these together we get L(x)=3x+6m/x

Setting the derivative of L(x) to zero we get:

0=3-6m/x^2 or 6m=3x^2 , 2m=x^2 so x=1414.2136ft

Using xy=3m we get y=2121.3203ft

But we still have to find L. L=3(1414.2136)+2(2121.3203) so that L=8485.2814ft. This is the shortest amount of fence he can use to construct this fence including the segment bisecting the middle of the area.