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 [math] A_1=xy [/math] and circumference be [math]L = 2x + 2y \Leftrightarrow y=\frac{L}{2} - x[/math] and
[math]A_1(x) = x ( \frac{L}{2} - x )[/math] has maximum value [math]A_1=\frac{L^2}{16}[/math] for [math] x = \frac{L}{4}[/math].
2) Also examine the case where the wall is used [math]L = 2x + y[/math].
[math]A_2(x) = x ( L-2x )[/math] has maximum value [math] A_2 = \frac{L^2}{8}[/math] for [math] x = \frac{L}{4}[/math]
3) The [b]circle[/b] has a greater area to circumference ratio than the [b]rectangle[/b] ( or [b]square[/b] ) :
ratio k = [math] \frac{A}{L^2} = \frac{\pi R^2}{ {2 \pi R}^2} = \frac{1}{4\pi} \approx [/math] 0,07957747 > 0,0675 = [math] \frac{1}{16} [/math]
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]
[math]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} [/math]
[math]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}[/math]
Differential calculus assuming that the derivative = 0 at maximum point yields
[math] \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}[/math]
[math] \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} [/math]
