A farmer wishes to fence off a rectangular pasture along the bank of a river. The area of the pasture is 3200 $yd^2$. There is no fencing needed along the river bank. Find the dimensions of the pasture that'll require the least amount of fencing. $$\\$$

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I'm not sure how to begin this problem.

I know the area is $lw$ = 3200.

Please help.

  • $\begingroup$ Try drawing a picture to help visualize what the question is asking. $\endgroup$ – BBot Jun 27 '17 at 22:50
  • $\begingroup$ I suppose we're to assume the river has a straight bank? It would make the problem more interesting if the bank was given by an arbitrary function, though. $\endgroup$ – Χpẘ Jun 27 '17 at 23:03
  • $\begingroup$ @Χpẘ, it wouldn't matter what the function is for the river bank because you don't need fencing around there $\endgroup$ – BBot Jun 27 '17 at 23:11
  • $\begingroup$ @Bbot Imagine the riverbank runs "horizontally" on a graph in the shape of a sine wave, $f(x)=a\cdot \sin(bx+\frac{3\pi}{2})+a$, for scaling constants $a,b$. The "northwest" corner of the pasture is at the origin. The west fence would be length $w$, but the east fence would $w+\sin(bl+\frac{3\pi}{2})$. You could draw this if it isn't clear. $\endgroup$ – Χpẘ Jun 27 '17 at 23:23
  • $\begingroup$ @Χpẘ, yes I agree with what you are saying but for this specific problem it doesn't state a function so just use the perimeter which doesn't include the river side. Thus, we use $x$+2$y$ $\endgroup$ – BBot Jun 27 '17 at 23:26

Since the pasture is a rectangle, the Area= $xy$= 3200.

The minimum fencing required is f($x$,$y$) = $x$+2$y$.$$\\$$ $fx$→ 1 = λy

$\implies$ $\frac{1}{y}$ = λ

$fy$→ 2 = λx

$\implies$ $\frac{2}{x}$ = λ

Set them equal and cross multiply: $\frac{1}{y}$=$\frac{2}{x}$

$\implies$ 2$y$=$x$ $$\\$$ Remember that the Area: $xy$= 3200 and we just found what $x$ equals.

$xy$= 3200

= (2$y$)$y$ = 3200

= 2$y^2$ = 3200

= $y^2$ = 1600

= $y$ = ±40

$y$ = 40 because you can't have a negative yard. $$\\$$

Remember that 2$y$=$x$ and we just found what $y$ equals.


= 2(40) = $x$

= $x$ = 80 $$\\$$ So your dimension are 80 x 40 yards.

  • $\begingroup$ how did you get $x$+2$y$? $\endgroup$ – AmR Jun 27 '17 at 23:12
  • $\begingroup$ I suggest you draw a visual $\endgroup$ – BBot Jun 27 '17 at 23:14
  • $\begingroup$ Make one side the river, one side the $x$, and the other two sides as $y$ $\endgroup$ – BBot Jun 27 '17 at 23:14
  • $\begingroup$ ohhh, okay. I get it. thank you! $\endgroup$ – AmR Jun 27 '17 at 23:20
  • $\begingroup$ Just for fun, I get the following if the curve of the river is given by $g(x)$, and $f(l,w)$ is the length of the fence: $f(l,w)=g(0)+w+g(l)+w$ and $\int_0^l g(x) dx + lw = 3200$, so $f$ in terms of $l$ is $f(l)=g(0)+g(l)+\frac{2(3200-\int_0^l g(x) dx)}{l}$. This gives the same result if $g(x)=0$ for all $x$ $\endgroup$ – Χpẘ Jun 27 '17 at 23:45

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