# Discovering relations between area and circumference.

Which is the optimal rectangular area obtainable with fencing material of length L =10 ? These kind of optimization problems appear with equations of 2nd degree at age 15-18, but this visual version can in my opinion also be examined at 11-14. The calculus-analysis would look like this. 1) Let area be $A_1=xy$ and circumference be $L = 2x + 2y \Leftrightarrow y=\frac{L}{2} - x$ and $A_1(x) = x ( \frac{L}{2} - x )$ has maximum value $A_1=\frac{L^2}{16}$ for $x = \frac{L}{4}$. 2) Also examine the case where the wall is used $L = 2x + y$. $A_2(x) = x ( L-2x )$ has maximum value $A_2 = \frac{L^2}{8}$ for $x = \frac{L}{4}$ 3) The [b]circle[/b] has a greater area to circumference ratio than the [b]rectangle[/b] ( or [b]square[/b] ) : ratio k = $\frac{A}{L^2} = \frac{\pi R^2}{ {2 \pi R}^2} = \frac{1}{4\pi} \approx$ 0,07957747 > 0,0675 = $\frac{1}{16}$ PS. Merci, thanks and credit to V. Launay for earlier development on Geogebra 3.0. [url]http://www.geogebra.org/en/upload/files/vlaunay/activite1.html[/url] $A_1(x)=-x^2+\frac{L}{2}x=-(x^2-\frac{L}{2}+\frac{L^2}{16})+\frac{L^2}{16}=-\left(x-\frac{L}{4}\right)^2 + \frac{L^2}{16}$ $A_2(x)=-2x^2+Lx=-2(x^2-\frac{L}{2}x+\frac{L^2}{16})+\frac{L^2}{8}=-2\left(x-\frac{L}{4}\right)^2+\frac{L^2}{8}$ Differential calculus assuming that the derivative = 0 at maximum point yields $\frac{dA_1}{dx} = - 2x + \frac{L}{2} = 0 \Leftrightarrow x = \frac{L}{4} \Leftrightarrow y_1 = \frac{L}{2} - \frac{L}{4} = \frac{L}{4} \Rightarrow A_1=xy_1 = \frac{L}{4} \cdot \frac{L}{4} = \frac{L^2}{16}$ $\frac{dA_2}{dx} = -4x + L = 0 \Leftrightarrow x = \frac{L}{4} \Leftrightarrow y_2 = L - 2 \frac{L}{4} = \frac{L}{2} \Rightarrow A_2=xy_2 = \frac{L}{4} \cdot \frac{L}{2} = \frac{L^2}{8}$

Material Type
Activity
Tags
area  circumference  parable  equation  second  degree
Target Group (Age)
15 – 18
Language
English (United Kingdom)

GeoGebra version
4.0
Views
3703

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