So the q is; You need to enclose a rectangular field with area 9600m^2, then divide the field into 2 sections. Whats the minimal amount of fencing required to accomplish this. So i have 3x+2y, and 9600=xy. i TRIED ISOLATING the second formula and plugging it into the original and finding the derivative. However I keep getting a negative answer. Someone please help
If f(x,y) is the amount of fence, then
f = 3x+2y = 3x+2(9600/x) = 3x + 19200/x
df/dx = 3 - 19200/x^2
df/dx=0 at x = ±80
f has a max at x = -80, and a min at x=80.
Too bad you didn't show your work ...
To find the minimal amount of fencing required to enclose a rectangular field with an area of 9600m^2 and divide it into two sections, you need to find the dimensions of the field that minimize the perimeter.
Let's start by using the given formulas:
1) Area formula: xy = 9600
2) Perimeter formula: 3x + 2y
To minimize the perimeter, we need to find the critical points of the function 3x + 2y. Let's isolate one variable in the area formula and substitute it into the perimeter formula to express the perimeter only in terms of one variable.
From the area formula: y = 9600/x
Substituting this into the perimeter formula, we get:
P = 3x + 2(9600/x)
Simplifying:
P = 3x + 19200/x
Now, to find the critical points, we need to find where the derivative of the perimeter function is equal to zero. Let's differentiate the perimeter function P with respect to x:
dP/dx = 3 - 19200/x^2
Setting dP/dx equal to zero:
3 - 19200/x^2 = 0
Rearranging and solving for x^2:
19200/x^2 = 3
x^2 = 19200/3
x^2 = 6400
x = ±80
Since the dimensions of a field cannot be negative, we take x = 80.
Now, substitute this value of x back into the area formula to find y:
xy = 9600
80y = 9600
y = 120
Therefore, the dimensions of the field that minimize the perimeter are x = 80m and y = 120m.
To find the minimal amount of fencing required, we can substitute these values into the perimeter formula:
P = 3x + 2y
P = 3(80) + 2(120)
P = 240 + 240
P = 480
So, the minimal amount of fencing required to enclose the field and divide it into two sections is 480 meters.