Respuesta :
Lets say that,
x = width of rectangle
y = length of rectangle
The area of the rectangle
is given as:
A = x * y = 2500000 ft^2
Rewriting in terms of y:
y = 2500000 / x
The length of fencing needed is equal to the perimeter of
rectangle plus the middle fence:
L = 2(x + y) + x = 3x + 2y
Substituting the value of
y:
L = 3x + 2(2500000/x) = 3x + 5*10^6*x^(-1)
The minima can be obtained by taking the 1st
derivative of the equation then equating dL/dx = 0 :
dL/dx = 3 - 5*10^6*x^(-2)
3 - 5*10^6*x^(-2) = 0
x^-2 = 6*10^-7
x = 1291
Calculating for y:
y = 2500000 / x = 2500000 / 1291
y = 1936.48
Therefore the shortest length of fence needed is:
L = 3x + 2y = 3(1291) + 2(1936.48)
L = 7746 ft
The shortest length of fence that the rancher can use [tex]\boxed{7746{\text{ feet}}}.[/tex]
Further explanation:
Given:
The area of the rectangular field is [tex]2500000{\text{ fee}}{{\text{t}}^2}.[/tex]
Explanation:
Consider the length of the rectangular field be “x”.
Consider the width of the box as “y”.
The area of the rectangular field can be expressed as follows,
[tex]\boxed{Area = \left( x \right) \times\left( y \right)}[/tex]
The given area of the field is [tex]2500000{\text{ fee}}{{\text{t}}^2}.[/tex]
[tex]\begin{aligned}{\text{Area}} &= 2500000\\\left( x \right) \times \left( y \right)&= 2500000\\y&=\frac{{2500000}}{x}\\\end{aligned}[/tex]
The perimeter of the rectangular field with a fence down in the middle parallel to one side can be expressed as follows,
[tex]\begin{aligned}{\text{Perimeter}}&= 3x + 2y\\&= 3x + 2 \times \frac{{2500000}}{x}\\&= 3x + \frac{{5000000}}{x}\\\end{aligned}[/tex]
Differentiate the above equation with respect to x.
[tex]\begin{aligned}\frac{d}{{dx}}\left( {{\text{Perimeter}}} \right) &= \frac{d}{{dx}}\left( {3x + \frac{{5000000}}{x}} \right)\\&= 3 - \frac{{5000000}}{{{x^2}}}\\\end{aligned}[/tex]
Again differentiate with respect to x.
[tex]\dfrac{{{d^2}}}{{d{x^2}}}\left( {{\text{Perimeter}}} \right) = \dfrac{{10000000}}{{{x^3}}}[/tex]
Substitute the first derivative equal to zero.
[tex]\begin{aligned}\frac{d}{{dx}}\left( {{\text{Perimeter}}} \right)&= 0\\3- \frac{{5000000}}{{{x^2}}}&= 0\\3&= \frac{{5000000}}{{{x^2}}}\\{x^2} &= \frac{{5000000}}{3}\\x&= \sqrt {\frac{{5000000}}{3}}\\x&= 1291\\\end{aligned}[/tex]
The value of y can be calculated as follows,
[tex]\begin{aligned}y&= \frac{{2500000}}{{1291}}\\&= 1936.50\\\end{aligned}[/tex]
The shortest length can be calculated as follows,
[tex]\begin{aligned}{\text{Length} &= 3x + 2y\\&= 3 \times 1291 + 2 \times 1936.50\\&= 3873 + 3873\\&= 7746{\text{ feet}}\\\end{aligned}[/tex]
The shortest length of fence that the rancher can use [tex]\boxed{7746{\text{ feet}}}.[/tex]
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Answer details:
Grade: High School
Subject: Mathematics
Chapter: Area
Keywords: square, area, fence, rancher, feet, rectangle field, divide, half, fence down, middle parallel, one side, shortest length, perimeter, circumference.
