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A three-sided fence is to be built next to a straight section of a river, which forms the fourth side of a rectangular region. The enclosed area is equal to 1800 ft^2. Find the minimum perimeter and the dimensions of the corresponding enclosure.

What I have so far: Area = length * width = 1800, Perimeter = 2 * length + 2 * width

Substituting gives Perimeter = 3600/width + 2 * width. The derivative is then -(3600)/(width^2) + 2 = 0. But this ultimately results into width^2 = -1800 and that's not possible??

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    $\begingroup$ One hint: perimeter is only 1* length + 2 * width, because on one side there is the river. $\endgroup$ – callculus Oct 3 '14 at 22:29
  • $\begingroup$ Intuitively, that rectangle has to be half a square, since, when given n sides, the regular polygon is the closest thing to a circle, the circle being the two dimensional geometric figure with optimal perimeter-to-area ratio. In three dimensions, it would be a sphere. Or, in your case, half a cube. $\endgroup$ – Lucian Oct 3 '14 at 23:21
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What got you into difficulty is that you did not take into account that one side of the rectangular region is bounded by the river. Thus, the perimeter of the fence includes just three sides of the rectangle.

The area of the rectangular region is $1800 = lw$, so

$$w = \frac{1800~\text{ft.}^2}{l}$$

The perimeter of the fence is $P = l + 2w$. Substituting for $w$ yields

\begin{align*} P(l) & = l + 2\left(\frac{1800~\text{ft.}^2}{l}\right)\\ & = l + \frac{3600~\text{ft.}^2}{l} \end{align*}

Differentiating yields

$$P'(l) = 1 - \frac{3600~\text{ft.}^2}{l^2}$$

Setting the derivative equal to zero yields $l = 60~\text{ft.}$ The derivative is negative if $l < 60~\text{ft.}$ and positive if $l > 60~\text{ft.}$ Thus, by the First Derivative Test, the perimeter is minimized if $l = 60~\text{ft.}$

If $l = 60~\text{ft.}$, then the width of the rectangular region is

$$w = \frac{1800~\text{ft.}^2}{l} = \frac{1800~\text{ft.}^2}{60~\text{ft.}} = 30~\text{ft.}$$

so the fence has perimeter

$$P = l + 2w = 60~\text{ft.} + 2(30~\text{ft.}) = 120~\text{ft.}$$

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The derivative is then -(3600)/(width^2) + 2 = 0. But this ultimately results into width^2 = -1800 and that's not possible?

Only one result of your equation is negative.

$-\frac{3600}{w^2}+2=0$

multiplying both sides by $w^2$

$-3600+2w^2=0$

adding 3600 on both sides

$2w^2=3600$

Dividing both sides by 2 and taking the square root on both sides.

$w_{1,2}=\pm \sqrt {1800}$

But as I commented, the equation is wrong.

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Take the same three sided enclosure and build it on the other side of the river. We get a hexagonal enclosure that the river divides evenly, of area $2\times 1800 \text{ft}^2= 3600 \text{ft}^2$. The perimeter doubles too. For minimum perimeter this hexagonal enclosure should be a regular hexagon of area ${ 3600 \text{ft}^2}$,

Hence the enclosure will be a half of a regular hexagon, with three sides of length $$20 \sqrt{2} \sqrt[4]{3} \simeq 37 \text{ft}\ 2\frac{11}{16} \text{in}$$

and a perimeter of $$60 \sqrt{2} \sqrt[4]{3} \simeq 111\ \text{ft}\ 8\frac{5}{64}\text{in}$$

This beats the rectangle enclosure of perimeter $120\text{ft}$.

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It is more comfortalbe if you use letters than whole words in your mathtematical expressions. So you have

$$A=l \cdot w=1800$$ $$p=2l+2w \tag{1}$$ and therefore $$p(w)=\frac{3600}{w}+2w$$ and setting the derivate to $0$ gives $$-\frac{3600}{w^2}+2=0 \tag{2}$$

You get $$ w^2=-1800$$ and therefore $$w=\pm \sqrt{-1800}=\pm 30\sqrt{2}\,i$$ This is a complex solution, but actually this is not a solution of $(2)$. If you substitut $w=\sqrt{-1800}$ in the LHS of $(2)$ you get $$-\frac{3600}{(\sqrt{-1800})^2}+2=-\frac{3600}{-1800}+2=-(-2)+2=4$$. This is not the RHS which is $0$. So you made an error when solving $(2)$

But you already made an error earlier whe you establieshed equation $(1)$. From the text I would conclude that one side of the square is bordered by the river and there is no fence on this side.

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All these answers are correct, but they all set the derivative of some function to zero, an understandable reflex when one is asked to find the minimum of some variable subject to constraints. However, a simpler route is available. Using the variables in the accepted answer of @N. F. Taussig, $$\begin{align*} P(l) & = l + 2\left(\frac{1800~\text{ft.}^2}{l}\right)\\ & = l + \frac{3600~\text{ft.}^2}{l} \end{align*}$$Replacing $P(l)$ with simply $P$, multiplying through by $l$, and re-arranging.we obtain$$l^2-Pl+3600 =0$$Applying the Quadratic Formula, we obtain:$$l = \frac{P\pm\sqrt{P^2-4\times 3600}}{2}$$For a real solution, the discriminant must be non-negative. $P=\sqrt{4 \times 3600}=120$ is the smallest possible value of P that keeps the discriminant non-negative...

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