Mathematics-Online problems: Solution to the problem of the (previous) week

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	Mathematics-Online problems:
	Solution to the problem of the (previous) week

Problem:

From a square piece of cardboard with an edge length of $10\ \mathrm{cm}$ a square pyramid should be made as is illustrated in the figure.

$\includegraphics[width=.3\linewidth]{TdM_12_A2_bild1}$ $\includegraphics[width=.4\linewidth]{TdM_12_A2_bild2}$

For which edge length does the volume become a maximum?
Also determine $V_{\max}$ as well as the surface area (base included) and the height of the optimal pyramid.

Answer:

     cm

     cm

$V_{\max} =$      cm $^{3}$

     cm $^{2}$

(The results should be correct to four decimal places.)

Solution:

Let

be the slant height of the pyramid. Then, we have

$\displaystyle 2s + a = 10\sqrt{2}$

and

$\displaystyle h = \sqrt{s^2 - (a/2)^2}\, .$

Substituting this expression into the formula of the pyramid's volume, we obtain

$\displaystyle V = \dfrac{1}{3}a^2 h = \dfrac{1}{3}a^2 \sqrt{\left(\dfrac{10\sqrt{2}-a}{2}\right)^2 - \left(\dfrac{a}{2}\right)^2 } \,.$

To simplify matters we maximize

$\displaystyle f(a) = 9 V^2 = a^4 \left(50 - 5\sqrt{2} a\right) \,.$

From

$\displaystyle 0 = f'(a) = 200 a^3 - 25\sqrt{2} a^4$

follows

$\displaystyle a = 4\sqrt{2} \approx 5.6569$

being the only positive solution.

Moreover,

$\displaystyle s = \dfrac{10\sqrt{2} - a}{2} = 3\sqrt{2}\,, %\approx 4.2426\,,$

$\displaystyle h = \sqrt{\left(3\sqrt{2}\right)^2 - \left(2\sqrt{2}\right)^2} = \sqrt{10} \approx 3.1623\,,$

$\displaystyle V_{\max} = \dfrac{1}{3} \cdot \left(4\sqrt{2}\right)^2 \cdot \sqrt{10} = \dfrac{32\sqrt{10}}3 \approx %0.0337 L^3 = 33.7310\,,$

$\displaystyle S = a^2 + 2as = 32 + 2\cdot 4\sqrt{2}\cdot 3\sqrt{2} %= \dfrac{4}{5}L^2 = 80\,. %= 0.8 L^2$