# Find the dimensions of the rectangle that will give the minimum perimeter.

a farmer wants to make a rectangular paddock with an area of $$4000 m^2$$ However, fencing costs are high and she wants the paddock to have a minimum perimeter.

I have found the perimeter:

$$x\cdot y = 4000\\ y = \frac{4000}{x}$$

\begin{align}\text{Perimeter} &= 2x + 2y\\ &= 2x + 2(4000/x)\\ &= 2x + (8000/x)\end{align}

How do I find the dimensions that will give the minimum perimeter?

You could just minimize the function that you found for the perimeter. The minimum is attained when the derivative is zero. So calculate: $$\frac{d}{dx} perimeter= \frac{d}{dx}(2x+\frac{8000}{x})=2-\frac{8000}{x^2}=0.$$ This gives $2=\frac{8000}{x^2}$, so $4000=x^2$. We then get $$x=\sqrt{4000},$$ since the negative solution is not an option.

• A rectangle has two sides longer than the other two. square root of 4000 gives me one answer. if breath and height are the same, it would be a square. Commented Nov 10, 2016 at 11:14
• A shape being a rectangle means that it has 90 degree angles, so a square is just a specific case of a rectangle. Commented Nov 10, 2016 at 11:29
• ok so both x and y are 63.2455...? Commented Nov 10, 2016 at 11:35
• That's true. Since y=4000/x=4000/\sqrt(4000)=\sqrt(4000) Commented Nov 10, 2016 at 11:48
• If fences are sold in integer length, so the size will be 64*63, how can I account this into the formula?
– Roy
Commented Oct 31, 2019 at 2:02

As $x>0$

by AM GM inequality $$\dfrac{2x+\dfrac{8000}x}2\ge\sqrt{2x\cdot\dfrac{8000}x}$$

which can also be written as

$$2x+\dfrac{8000}x=\left(\sqrt{2x}-\sqrt{\dfrac{8000}x}\right)^2+2\sqrt{2x\cdot\dfrac{8000}x}\ge2\sqrt{2x\cdot\dfrac{8000}x}$$

• @Utsav, Find the last part. Do you know that for real $a, a^2\ge0$ Commented Nov 10, 2016 at 9:12

Rectangle with maximal area and minimal perimeter is square if $x\cdot y=4000$ then $x=y$ so $x^2=4000$ and $x=\sqrt{4000}=63,2455...$

• I suspect that the OP wants to treat this with the method of Lagrange multipliers. Commented Nov 10, 2016 at 9:56

$$2x+2y=p,\, y= p/2 -x ; \, A= y\cdot x = (p/2 -x)\cdot x$$

$$\frac{dA}{dx} =0 \rightarrow x = p/4 =y ,\quad A= p^2/16.$$

For maximum area it should be a square of quarter perimeter length as side.

If you use $$A.M \geq G.M$$, you get $$2x + 2y \geq 4\sqrt{xy} = 80\sqrt{10}$$ and equality holds iff $$x=y$$.

To minimise $$2(x+y)$$ subject to $$xy=A$$ note that: $$(x+y)^2=4xy+(x-y)^2=4A+(x-y)^2$$

Now since $$x$$ and $$y$$ are both positive, the minimum value of $$2(x+y)$$ occurs when $$4(x+y)^2$$ is a minimum and hence when $$(x+y)^2$$ is a minimum.

Since $$(x-y)^2$$ is non-negative, the minimum occurs when $$x-y=0$$ ie when you have a square and $$x+y=2\sqrt A$$ and $$2(x+y)=4\sqrt A$$