Answer:
Step-by-step explanation:
The the area of the rectangular pen be A = LW
L is the length
W is the width
Given
A = 3136ft²
3136 = LW..........1
Given the perimeter P = 2L+2W....... 2
From 1; L = 3136/W
Substitute into 2
P = 2(3136/W)+2W
P = 6272/W + 2W
In order to minimize the amount of material needed, then dP/dW = 0
dP/dW = -6272/W² + 2
0 = -6272/W² + 2
6272/W² = 2
cross multiply
6272 = 2W²
W² = 3136
W = √3136
W = 56ft
Since A = LW
3136 = 56L
L = 3136/56
L = 56ft
Hence the dimensions of the rectangular pen that minimize the amount of material needed is 56ft by 56ft
The dimensions of the rectangle are needed.
The minimum dimensions are length 56 feet and width 56 feet.
Let [tex]x[/tex] be the length of rectangle
[tex]y[/tex] be the width of rectangle
The area of the rectangle is
[tex]A=3136\\\Rightarrow xy=3136\\\Rightarrow y=\dfrac{3136}{x}[/tex]
Perimeter is given by
[tex]P=2(x+y)\\\Rightarrow P=2(x+\dfrac{3136}{x})[/tex]
Differentiating with respect to [tex]x[/tex]
[tex]P'=2(1-\dfrac{3136}{x^2})[/tex]
Equating with zero
[tex]0=2(1-\dfrac{3136}{x^2})\\\Rightarrow x=\sqrt{3136}\\\Rightarrow x=56[/tex]
Double derivative of perimeter at [tex]x=56[/tex]
[tex]P''=2\times 2(\dfrac{3136}{x^3})\\\Rightarrow P''(56)=2\times 2(\dfrac{3136}{56^3})=0.07>0[/tex]
Since, it is greater than zero the perimeter will be minimum at [tex]x=56[/tex]
Finding [tex]y[/tex]
[tex]y=\dfrac{3136}{x}\\\Rightarrow y=\dfrac{3136}{56}\\\Rightarrow y=56[/tex]
The minimum dimensions are length 56 feet and width 56 feet.
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