geometry
asks;
Determine the dimensions of a square based prism box with each volume that requires the least material to make.
$$a) 512 \ \ cm^3$$
$$b) 1000\ \ cm$3$$
$$c) 750\ \ cm^3$$
I attempted the question, sort of understand it. if someone can show me how to do one of them, I can most likely do the rest on my own .
How absurd is that 4 days later the answer just struck me out of nowhere. Basically, my approach to the whole thing was wrong. I thought that since what's cut off of the triangular prism is amorphous thus what's left from it (the body which volume I'm interested in) is also amorphous, but this is not true.
If we have a sphere what's cut by a triangular prism which edges tangent to the sphere's surface (just like in my sketches), so basically doing the same as before, but vice versa throw new light upon the whole problem. This way what was cut off is 2 “spherical cap” (and of course, the half of the whole sphere, which also a spherical cap by the way), which volume is easy to calculate!
Now let's talk about the solution, shall we? First, what should we know about the spherical cap? The most important part is that we know its volume! If there's a spherical cap of a sphere with radius $R$ (the diameter is $a$), which height is $h$ and the radius of its base is $r$, then its volume is $V_{cap}=\frac{\pi h}{6}(3r^2+h^2)=\frac{\pi h^2}{3}(3R-h)$. We know that $R=\frac{a}{2}$, $r=\frac{\sqrt{2}}{2}R=\frac{\sqrt{2}}{4}a$ and $h=R-\sqrt{R^2-r^2}$ $=\frac{2-\sqrt{2}}{2}R$ $=\frac{2-\sqrt{2}}{4}a$, so we can express the volume as a function of $a$, as I originally wanted! The result is $V_{cap}=\frac{8-5\sqrt{2}}{96}\pi a^3=\frac{8-5\sqrt{2}}{12}\pi R^3$.
We're out of the woods now, because the only thing left to do is to subtract twice of that from the volume of the hemisphere, which means $V=\frac{1}{2}V_\bigcirc-2V_{cap}=\frac{1}{12}\pi a^3-\frac{8-5\sqrt{2}}{48}\pi a^3=\frac{5\sqrt{2}-4}{48}\pi a^3$ $=\frac{5\sqrt{2}-4}{6}\pi R^3$ and which means it's $\frac{5\sqrt{2}-4}{24}\pi V_\Box \approx \frac{3}{24}\pi V_\Box \neq \frac{1}{12}\pi V_\Box$. So, that's it! This was what I wanted to know: the formula of the volume of that body as a function of the sphere's diameter.
Suppose a square of side $s\le\frac12$ is placed randomly inside a unit square, without rotation. The probability that the $x$-coordinate of the center of the unit square is within the $x$-span of the smaller square is $\frac{s}{1-s}$. That is, when covering the center of the unit square, the left side of the smaller square can be $s$ to $0$ units to the left of the center of the unit square, while, without restriction, the left side of the smaller square can go from $1$ to $s$ units to the left of the right side of the unit square. The same is true for the $y$-coordinate. So the probability that the center of the larger square is within the smaller square is $$ \left(\frac{s}{1-s}\right)^2\tag1 $$ In your question, $s=\frac1{\sqrt{10}}$, which gives a probability of $$ \bbox[5px,border:2px solid #C0A000]{\frac1{11-2\sqrt{10}}\approx0.213883399}\tag2 $$
Geometry and Probability
The smaller square rotated by $0\le\theta\le\frac\pi4$ will span a width and height of $$ d=s(\cos(\theta)+\sin(\theta))\tag3 $$ Note that $(3)$ says that $d:\left[0,\frac\pi4\right]\mapsto\left[s,s\sqrt2\right]$ monotonically.
If the smaller square is to stay inside the unit square, its center is constrained to a non-rotated square of side $1-d$ concentric with the unit square (shown below as a red square). If the center of the unit square is to be inside the smaller square, the center of the smaller square needs to be inside a square with the same size and rotation, but concentric with the unit square (shown below as a green square).
Thus, the locus of the center of the smaller square so that the smaller square is contained in the unit square, and so that the smaller square contains the center of the unit square is the intersection of the red and green squares (shown below in purple).
For a given size and rotation of the smaller square, given that the smaller square is contained in the unit square, the probability that the center of the unit square is inside the smaller square is the ratio of the area of the purple region divided by the area of the red square.
However, to get the proper weighting of the probabilities for the different angles, we need to weight the probabilities by the area of the red square since that is the probability that the rotated square is contained in the unit square. This means we compute the integral of the area of the purple region and divide that by the integral of the area of the red region.
Computation of the Areas
The area of the red square is $$ \begin{align} \text{area}_\text{red} &=(1-d)^2\tag{4a}\\[6pt] &=(1-s(\sin(\theta)+\cos(\theta)))^2\tag{4b}\\[6pt] &=1-2s(\sin(\theta)+\cos(\theta))+s^2(1+2\sin(\theta)\cos(\theta))\tag{4c}\\ \int_0^{\pi/4}\text{area}_\text{red}\,\mathrm{d}\theta &=\frac\pi4-2s+\left(\frac\pi4+\frac12\right)s^2\tag{4d} \end{align} $$ If $d\le\frac12$, then, because $d\lt1-d$, the green square is completely contained within the red square and so the area of the purple region is $$ \begin{align} \text{area}_\text{purple} &=s^2\tag{5a}\\ \int\text{area}_\text{purple}\,\mathrm{d}\theta &=\theta s^2\tag{5b}\\ &=a_1(s,\theta)\tag{5c} \end{align} $$ We leave $\text{(5b)}$ indefinite since we may need to restrict $\theta$ due to the condition $d\le\frac12$.
For $d\gt\frac12$, we will refer to the following diagram:
Since the side length of the red square $1-d$ and the height of the rotated green square is $d$, the four corners of the green square that extend beyond the red square have altitude $\frac{d-(1-d)}2=d-\frac12$ and base $\frac{d-1/2}{\sin(\theta)\cos(\theta)}$. If the bases of these triangles are smaller than the sides of the red square, then $\frac{d-1/2}{\sin(\theta)\cos(\theta)}\lt1-d$, which is equivalent to $d\lt\frac{1+\sin(2\theta)}{2+\sin(2\theta)}$.
Thus, if $\frac12\lt d\lt\frac{1+\sin(2\theta)}{2+\sin(2\theta)}$, the area of the purple region is $$ \begin{align} \text{area}_\text{purple} &=s^2-\frac{2(d-1/2)^2}{\sin(\theta)\cos(\theta)}\tag{6a}\\ &=s^2-\frac{2\left(s(\sin(\theta)+\cos(\theta))-\frac12\right)^2}{\sin(\theta)\cos(\theta)}\tag{6b}\\[3pt] &=\scriptsize-\frac12\csc(\theta)\sec(\theta)+2(\csc(\theta)+\sec(\theta))s-(3+2\csc(\theta)\sec(\theta))s^2\tag{6c}\\[6pt] \int\text{area}_\text{purple}\,\mathrm{d}\theta &=\scriptsize-\frac12\log(\tan(\theta))+2\log\left(\tan\left(\tfrac\theta2\right)\tan\left(\tfrac\pi4{+}\tfrac\theta2\right)\right)s-\left(3\theta+2\log(\tan(\theta))\right)s^2\tag{6d}\\[9pt] &=a_2(s,\theta)\tag{6e} \end{align} $$ If $d\ge\frac{1+\sin(2\theta)}{2+\sin(2\theta)}$, the red square is completely contained within the green square and so the area of the purple region is the same as the area of the red square: $$ \begin{align} \text{area}_\text{purple} &=(1-d)^2\tag{7a}\\[3pt] &=(1-s(\sin(\theta)+\cos(\theta)))^2\tag{7b}\\[3pt] &=1-2(\sin(\theta)+\cos(\theta))s+(1+2\sin(\theta)\cos(\theta))s^2\tag{7c}\\ \int\text{area}_\text{purple}\,\mathrm{d}\theta &=\theta-2(\sin(\theta)-\cos(\theta))s+\left(\theta+\sin^2(\theta)\right)s^2\tag{7d}\\ &=a_3(s,\theta)\tag{7e} \end{align} $$ Conditional Ranges
We need to apply each of $a_1$, $a_2$, or $a_3$ over particular parts of $\left[0,\frac\pi4\right]$. That is, we use $a_1$ over the green region of the chart below, $a_2$ over the yellow region, and $a_3$ over the red region.
To simplify the usage of the information in the chart above, define $$ \theta_1(s)=\frac12\sin^{-1}\left(\frac{1-4s^2}{4s^2}\right)\tag8 $$ and $$ \theta_2(s)=\frac12\sin^{-1}\left(\frac{1-4s^2+\sqrt{1-4s^2}}{2s^2}\right)\tag9 $$
If $s\le\frac1{2\sqrt2}$, we use $a_1$ for all $\theta\in\left[0,\frac\pi4\right]$. That is, the probability of the randomly rotated and positioned smaller square containing the center of the unit square is $$ \frac{a_1\!\left(s,\frac\pi4\right)-a_1\!\left(s,0\right)}{a_3\!\left(s,\frac\pi4\right)-a_3\!\left(s,0\right)}=\frac{\frac\pi4s^2}{\frac\pi4-2s+\left(\frac\pi4+\frac12\right)s^2}\tag{10} $$ If $\frac1{2\sqrt2}\lt s\le\frac{\sqrt2}3$, we use $a_1$ over $\theta\in[0,\theta_1(s)]$ and $a_2$ over $\theta\in\left[\theta_1(s),\frac\pi4\right]$. That is, the probability of the randomly rotated and positioned smaller square containing the center of the unit square is $$ \frac{a_2\!\left(s,\frac\pi4\right)-a_2(s,\theta_1(s))+a_1(s,\theta_1(s))-a_1(s,0)} {a_3\!\left(s,\frac\pi4\right)-a_3\!\left(s,0\right)}\tag{11} $$ If $\frac{\sqrt2}3\lt s\le\frac12$, we use $a_1$ over $\theta\in[0,\theta_1(s)]$, $a_2$ over $\theta\in[\theta_1(s),\theta_2(s)]$, and $a_3$ over $\theta\in\left[\theta_2(s),\frac\pi4\right]$. That is, the probability of the randomly rotated and positioned smaller square containing the center of the unit square is $$ \frac{a_3\!\left(s,\frac\pi4\right)-a_3(s,\theta_2(s))+a_2(s,\theta_2(s))-a_2(s,\theta_1(s))+a_1(s,\theta_1(s))-a_1(s,0)} {a_3\!\left(s,\frac\pi4\right)-a_3\!\left(s,0\right)}\tag{12} $$ In the question, $s=\frac1{\sqrt{10}}$, so we apply $(10)$ and get a probability of $$ \bbox[5px,border:2px solid #C0A000]{\frac\pi{11\pi+2-8\sqrt{10}}\approx0.279022074}\tag{13} $$ Here is a plot of the probability that a randomly rotated and positioned square of side $s$ inside the unit square contains the origin of the unit square:
Simulation
I ran a simulation placing the center of the smaller square, with side $s$, uniformly in a square, concentric with the unit square, and with side $1-s$, and rotating uniformly with $\theta\in\left[0,\frac\pi4\right]$.
All trials where the translated and rotated square did not fit completely in the unit square were rejected. The remaining trials were tallied counting those where the center of the unit square fell within the smaller square.
The results of $10,\!000,\!000$ trials with $s=\frac1{\sqrt{10}}$ were
non-rejected trials: $7,\!664,\!102$
successful trials: $2,\!136,\!729$
ratio: $0.278797046$
This is close to $0.279022074$.
Here is the Mathematica code for the simulation:
test[s_, x_, y_, \[Theta]_] :=
Module[{c = {x - 1/2, y - 1/2} (1 - s),
m = s {{Cos[\[Theta]], Sin[\[Theta]]}, {-Sin[\[Theta]], Cos[\[Theta]]}},
u = 1/2 {{-1, 1}, {1, 1}, {1, -1}, {-1, -1}},
cross = (#1.{{0, 1}, {-1, 0}}.#2) &, ul, ur, lr, ll},
{ul, ur, lr, ll} = u.m + {c, c, c, c};
If[Max[Abs[ul], Abs[ur], Abs[lr], Abs[ll]] > 1/2, {0, 0},
If[Min[cross[ur, ul], cross[ul, ll], cross[ll, lr], cross[lr, ur]] < 0,
{0, 1}, {1, 1}]]]
Module[{s = 1/Sqrt[10], n = 10000000, data = {0, 0}},
Do[data += test[s, RandomReal[1], RandomReal[1], RandomReal[Pi/4]], n];
data]
Best Answer
For a calculus based solution,
HINT:
Let the base square have side $x$, and the height be $y$.
The volume $V = x^2y$, giving $x = \sqrt{\frac{V}{y}}$
The surface area $A = 2x^2 + 4xy = 2\frac{V}{y} + 4\sqrt V \sqrt y$
Now you need to minimise $A(y) = 2\frac{V}{y} + 4\sqrt V \sqrt y$. Can you proceed from here? Find the general form for the minimal surface area in terms of the volume, then compute it for each case.