Calculus hw?

Question

A rancher wants to fence in an area of 3,200,000 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?

Answer 1

Call a is the length and b is the width

ab=3200000 and 2a + 3b = minimum

a = 3200000/b

6400000/b + 3b = min
3b^2 - b min + 6400000 = 0

delta = min^2 - 76800000
--> min=8746.4

--> b=1457.7 --> a=2195.2

minimum length of the fence = 2a+3b = 8763.6ft

Answer 2

let's set the width of the rectangle as x and the length of the rectangle as y, and the length of the fence that divide the field can be x or y, but in this problem, it doesn't really matter, x and y is interchangable, so we set the middle fence's length as x, so we have the area equaion xy = 3.2x10^6 , the total lengh of fence as L, so we have L = 3x + 2y. now put the area equaion in the total length equation. Since x = (3.2x10^6)/y, we get L = 3(3.2x10^6)/y + 2y , differentiate L . L' = 2 - 3(3.2x10^6)/y^2. When L' = 0, y = squareroot of (4.8x10^6), this would be the minimum value of y, and you can easily calculate the value of x, then put x and y in the function L, and you get the minimum value of L.

Answer 3

you must find 2 equations

area equation
xy=3200000

and one for the fence. you should know what to do from here

Answer 4

let x = width
3200000/x = lenght
L = 3x + 2l = 3x + 2(3200000/x)= 3x + 6400000/x
L' = 2x +(6400000)(-1x^(-2)) = 0
2x - 3200000/x^2 = 0
multiply by x^2
2x^3 = 3200000
x^3 = 1600000
x= 116.96 ft
l = 27360 ft
L = 3(116.96) + 2(27360) = 55071 ft