I have this question: A farmer plans to enclose a rectangular pasture adjacent to a river. The pasture must contain $320,000$ square meters in order to provide enough grass for the herd. What dimensions will require the least amount of fencing if no fencing is needed along the river? $$x = m$$ $$y = m$$

That what I did not sure what I'm doing wrong? $$xy=320000$$ $$2(x+y)=P$$ $$x+\frac{320,000}{x}=\frac{P}{Z}$$ $$1-\frac{320,000}{x^2}=0$$ $$x=400$$ $$y=400$$ My answer is wrong. I think $X$ needs to be larger then $Y$.

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    $\begingroup$ The "see figure" part was either a joke or else you copy too literally from somewhere...BTW, without a diagram I'm afraid your question is very hard to understand. $\endgroup$ – DonAntonio Jun 29 '13 at 19:05
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    $\begingroup$ Is Ho the farmer? $\endgroup$ – Chris Eagle Jun 29 '13 at 19:06
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    $\begingroup$ If the river requires no fence, the perimeter is probably not $2(x+y)$. Fix that and you should be okay. $\endgroup$ – John Douma Jun 29 '13 at 19:07

You set up the problem correctly, though we want for $2x + y = P$

$$ 2(x+y)= 2x + y = P$$ $$P = 2x+\dfrac{320,000}{x} = \frac{2x^2 + 320000}{x}$$

We want to minimize the perimeter needed, so we need to find the derivative of $P(x)$ and set that equal to zero, then solve for the zeros: the $x$ values that solve $P'(x) = 0$. Using the quotient rule:

$$P'(x) = \dfrac{4x\cdot x - (2x^2 + 320000)}{x^2} $$

Simplify and set equal to zero:

$$P'(x) = \dfrac{2x^2 - 320000}{x^2} = 0$$

$P'(x)$ will equal zero when the numerator is equal to zero. We certainly don't want a rectangle where $x = 0$, so we can affirm $x \neq 0 $ and hence the denominator will not be zero.

So, we simply solve for $x$: $$2x^2 - 320000 = 0 \iff x^2 - 160000 = (x + 400)(x - 400) = 0$$

So $$x = 400, y = \dfrac{320000}{400} = 800.$$


The length of the fence (which is not the whole perimeter) is $2x+y$ (assuming I have the orientation of the axes right). Note that $xy$ is fixed and you want to minimise $2x+y$. We note that $$(2x+y)^2=(2x-y)^2+8xy$$

With $xy$ fixed, the left hand side of this expression is a minimum when $2x=y$.


Another way to solve without calculus: For a fence-length $P$, area $A$, and dimensions $x$ and $y$ ($y$ parallel to river):$$x\times y = A$$ $$2x+y=P$$ Re-arranging the second equation to eliminate $y$ from the first:$$x\times (P-2x)=A$$ This re-arranges to the quadratic$$2x^2-Px+A=0 $$ with solutions:$$\frac{P ± \sqrt{ P^2-8A} }{4}$$The smallest value of P that yields a real root is $$P=\sqrt{8A}$$ and thus $$x=\frac{\sqrt{8A}}{4}=400$$


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