Respuesta :
Answer:
The shortest total length of fence with which this can be done is 40m^2
Step-by-step explanation:
Lets say that your area has width  [tex]w[/tex]   and length [tex]l[/tex].   Since you want to divided it into 3 equal areas by fences there would be 4  lengths and the area and perimeter would be given by the following equations.
[tex]3200 = w*l[/tex]
Perimeter = [tex]P = 4l+2w = 4l + \frac{2*3200}{l}[/tex]
We want to optimize the perimeter so
[tex]P'(l) = 4-\frac{6400}{l^2} = 0[/tex]
Solving for  [tex]l[/tex]   we get
[tex]l = 40.[/tex]
Since  [tex]3200 = l*w[/tex]   and  [tex]l = 40[/tex]   then   [tex]w = 3200/40 = 80[/tex]
The shortest length of the fence that can be done is 320 meters
Let the width of each partition be x, and the height of the fence be y.
So, the area of the fence is:
[tex]\mathbf{A = xy}[/tex]
Substitute 3200 for A
[tex]\mathbf{xy = 3200}[/tex]
The perimeter of the fence is calculated as:
[tex]\mathbf{P = 4x + 2y}[/tex]
Make x the subject in [tex]\mathbf{xy = 3200}[/tex]
[tex]\mathbf{x = \frac{3200}y}[/tex]
Substitute [tex]\mathbf{x = \frac{3200}y}[/tex] in [tex]\mathbf{P = 4x + 2y}[/tex]
[tex]\mathbf{P = 4\times \frac{3200}y + 2y}[/tex]
[tex]\mathbf{P = \frac{12800}y + 2y}[/tex]
Differentiate
[tex]\mathbf{P' = -\frac{12800}{y^2} + 2}[/tex]
Set to 0
[tex]\mathbf{ -\frac{12800}{y^2} + 2 = 0}[/tex]
Multiply through by y^2
[tex]\mathbf{ -12800 + 2y^2 = 0}[/tex]
Divide through by 2
[tex]\mathbf{ -6400 + y^2 = 0}[/tex]
So, we have:
[tex]\mathbf{ y^2=6400}[/tex]
Take square roots
[tex]\mathbf{ y=80}[/tex]
Substitute 80 for y in [tex]\mathbf{P = \frac{12800}y + 2y}[/tex]
[tex]\mathbf{P = \frac{12800}{80} + 2 \times 80}[/tex]
[tex]\mathbf{P = 160+ 160}[/tex]
[tex]\mathbf{P = 320}[/tex]
Hence, the shortest length of the fence is 320 meters
Read more about areas and perimeters at:
https://brainly.com/question/7154920
