A rancher wishes to create 4000 square meter rectangle enclosure. Two fences parallel to two of the sides will be used to create three equal areas. Find the dimensions of the fence that will require the least amount of fencing? Please help

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A rancher wishes to create 4000 square meter rectangle enclosure. Two fences parallel to two of the sides will be used to create three equal areas. Find the dimensions of the fence that will require the least amount of fencing? Please help

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Answer 1 · 2018-05-20T22:19:42

Length of fencing required = $80 \sqrt{20}$ $80sqrt20$ [mtrs]

Explanation:

Let the length of the large rectangle enclosing the other three rectangles have dimensions $x$ $x$ and $y$ $y$

Therefore $x y$ $x y$ = $4000$ $4000$ ....... $[1]$ $[1]$ The total length of the fencing must be the length enclosing the large rectangle i.e, $2 x$ $2x$ + $2 y$ $2y$ + the length of two cross pieces whose length is $x$ $x$ [ choosing to divide the large rectangle along it's $y$ $y$ length]

Therefore total length of fencing is $4 x + 2 y$ $4x+2y$ ......... $[2]$ $[2]$ . From ..... $[1]$ $[1]$ , $y = \frac{4000}{x}$ $y=4000/x$ and substituting this value of $x$ $x$ into....... $[2]$ $[2]$ will give us , length of fencing, say $L$ $L$ = $4 x + [2] [\frac{4000}{x}]$ $4x+[2][4000/x]$ Simplifying, $L = x + \frac{2000}{x}$ $L=x+2000/x$ , we need to differentiate this expression and set the derivitive = $0$ $0$ to find max/min.

$\frac{d L}{d x} = 1 - \frac{2000}{x^{2}}$ $[dL]/dx=1 - 2000/x^2$ = $0$ $0$ , which gives $x^{2} = 2000$ $x^2=2000$ , and simpilfying, $x = 10 \sqrt{20}$ $x=10sqrt20$ [ ignoring the the negative square root for $x$ $x$ , since we cannot have a negative length] ]

substituting this value of $x$ $x$ into...... $[1]$ $[1]$ will give a $y$ $y$ value of $20 \sqrt{20}$ $20sqrt20$

Therefore, from...... $[2]$ $[2]$ total length of fencing $L$ $L$ = $4 x + 2 y$ $4x+2y$ = $4 [10 \sqrt{20}] + 2 [20 \sqrt{20}$ $4[10sqrt20]+ 2[20sqrt20$ = $80 \sqrt{20}$ $80sqrt20$ '

To find if the value of $x$ $x$ minimises the the length of fencing to enclose the area of the rectangle we must check the second derivative, which if positive, will indicate a local minimum of the function $L$ $L$

$d \frac{L}{d x} = 1 - \frac{2000}{x^{2}}$ $dL/dx=1-2000/x^2$ , so $\frac{d}{d x} [d \frac{L}{d x}]$ $d/dx[dL/dx]$ = $\frac{2000}{x^{3}}$ $2000/x^3$ which is positive when $x$ $x$ = $10 \sqrt{20}$ $10sqrt20$ and so this value of $x$ $x$ will minimise $L$ $L$ . Hope this was helpful.

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A rancher wishes to create 4000 square meter rectangle enclosure. Two fences parallel to two of the sides will be used to create three equal areas. Find the dimensions of the fence that will require the least amount of fencing? Please help

1 Answer

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