# Prequel to “A Relentless Pursuit”

Fig. 1

Suppose we have two circular hoops of unit radius, centered on a common x-axis and a distance $2a$ apart. Suppose too, that a soap films extends between the two hoops, taking the form of a surface of revolution about the x-axis (see Fig. 2). Then if gravity is negligible the film takes up a state of stable, equilibrium in which its surface area is a minimum.

Fig. 2

Our problem is to find the function $y(x)$, satisfying the boundary conditions

$y(-a) = y(a) = 1,\quad\quad\quad(1)$

which makes the surface area

$A=2\pi\displaystyle\int\limits_{-a}^{a}y\sqrt{1+(y')^2}\;dx\quad\quad\quad(2)$

a minimum.

Let

$F(x,y, y') = 2\pi y \sqrt{1+(y')^2}.$

We have

$\frac{\partial F}{\partial y} = 2\pi \sqrt{1+(y')^2}$

and

$\frac{\partial F}{\partial y'} = 2 \pi y \cdot\frac{1}{2}\left(1+(y')^2\right)^{-\frac{1}{2}}\cdot 2y'=\frac{2 \pi y y'}{\sqrt{1+(y')^2}}.$

The Euler-Lagrange equation

$\frac{\partial F}{\partial y} - \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) = 0$

becomes

$\sqrt{1+(y')^2} - \frac{d}{dx}\left(\frac{y y'}{\sqrt{1+(y')^2}}\right) = 0.$

Fig. 3

Using Omega CAS Explorer (see Fig. 3), it can be simplified to:

$y \frac{d^2 y}{dx^2}- \left(\frac{dy}{dx}\right)^2=1.$

This is the differential equation solved in “A Relentless Pursuit” whose solution is

$y = C_1\cdot \cosh(\frac{x+C_2}{C_1}).$

We must then find $C_1$ and $C_2$ subject to the boundary condition (1), i.e.,

$C_1\cdot \cosh(\frac{a+C_2}{C_1}) = C_1\cdot\cosh(\frac{-a+C_2}{C_1})\implies \cosh(\frac{a+C_2}{C_1}) = \cosh(\frac{-a+C_2}{C_1}).$

The fact that $\cosh$ is an even function implies either

$a+C_2 = -a+C_2\quad\quad\quad(3)$

or

$a+C_2 = -(-a+C_2).\quad\quad\quad(4)$

While (3) is clearly false since it claims for all $a >0, a = -a$, (4) gives

$C_2=0.$

And so, the solution to boundary-value problem

$\begin{cases} y \frac{d^2 y}{dx^2}- \left(\frac{dy}{dx}\right)^2=1,\\ y(-a)=y(a)=1\end{cases}\quad\quad\quad(5)$

is

$y = C_1\cdot \cosh(\frac{x}{C_1}).\quad\quad\quad(6)$

To determine $C_1$, we deduce the following equation from the boundary conditions that $y=1$ at $x=\pm a:$

$C_1\cdot \cosh(\frac{a}{C_1}) = 1.\quad\quad\quad(7)$

This is a transcendental equation for $C_1$ that can not be solved explicitly. Nonetheless, we can examine it qualitatively.

Let

$\mu = \frac{a}{C_1}$

and express (7) as

$\cosh(\mu) = \frac{\mu}{a}.\quad\quad\quad(8)$

Fig. 4

A plot of (8)’s two sides in Fig. 4 shows that for sufficient small $a$, the curves $z = \cosh(\mu)$ and $z = \frac{\mu}{a}$ will intersect. However, as $a$ increases, $z=\frac{\mu}{a}$, a line whose slope is $\frac{1}{a}$ rotates clockwise towards $\mu$-axis. The curves will not intersect if $a$ is too large. The critical case is when $a=a^*$, the curves touch at a single point, so that

$\cosh(\mu) = \frac{\mu}{a^*}\quad\quad\quad(9)$

and $y=\frac{\mu}{a}$ is the tangent line of $z=\cosh(\mu),$ i.e.,

$\sinh(\mu) = \frac{1}{a^*}.\quad\quad\quad(10)$

Dividing (9) by (10) yields

$\coth(\mu) = \mu. \quad\quad\quad(11)$

What the mathematical model (5) predicts then is, as we gradually move the rings apart, the soap film breaks when the distance between the two rings reaches $2a^*$, and for $a > a^*$, there is no more soap film surface connects the two rings. This is confirmed by an experiment (see Fig. 1).

We compute the value of $a^*$, the maximum value of $a$ that supports a minimum area soap film surface as follows.

Fig. 5

Solving (11) for $\mu$ numerically (see Fig. 5), we obtain

$\mu = 1.1997.$

By (10), the corresponding value of

$a^* = \frac{1}{\sinh(\mu)} = \frac{1}{\sinh(1.1997)} = 0.6627$.

We also compute the surface area of the soap film from (2) and (6) (see Fig. 6). Namely,

$A = 2\pi \displaystyle\int\limits_{-a}^{a} C_1 \cosh\left(\frac{x}{C_1}\right) \sqrt{1+\left(\frac{d}{dx}C_1\cosh\left(\frac{x}{C_1}\right)\right)^2}\;dx = \pi C_1^2\left(\sinh\left(\frac{2a}{C_1}\right) + \frac{2a}{C_1}\right).$

Fig. 6

Exercise-1 Given $a=\frac{1}{2}$, solve (7) numerically for $C_1.$

Exercise-2 Without using a CAS, find the surface area of the soap film from (2) and (6).

# Jump!

Problem Given

$f(x) = e^x + \int\limits_{0}^{x} (t-x)f(t)\;dt\quad\quad\quad(\star)$

where $f(x)$ is a continuous function, find $f(x).$

Solution

From $(\star)$, we see that

$f(0) = 1;$

$f(x) = e^x + \int\limits_{0}^{x} t\cdot f(t) - x\cdot f(t) \;dt = e^x + \int\limits_{0}^{x} t\cdot f(t)\;dt-x\cdot \int\limits_{0}^{x}f(t)\;dt$.

And so,

$\frac{df(x)}{dx}=\frac{de^x}{dx} + \frac{d}{dx}\int\limits_{0}^{x}tf(t)\;dt - \frac{d}{dx}\left(x\cdot \int\limits_{0}^{x}f(t)\;dt\right)$

$=e^x+\frac{d}{dx}\int\limits_{0}^{x}tf(t)\;dt-\left(\int\limits_{0}^{x}f(t)\;dt + x\frac{d}{dx}\int\limits_{0}^{x}f(t)\;dt\right)$

$\overset{\textbf{FTC}}{=}e^x + xf(x) -\int\limits_{0}^{x} f(t)\;dt - xf(x)$

$= e^x - \int\limits_{0}^{x}f(t)\;dt$

That is,

$\frac{d}{dx}f(x)= e^x - \int\limits_{0}^{x}f(t)\;dt\implies f'(0) = 1.$

It follows that

$\frac{d^2f(x)}{dx^2}=\frac{d}{dx}\left(e^x-\int\limits_{0}^{x}f(t)\;dt\right)=e^x-\frac{d}{dx}\left(\int\limits_{0}^{x}f(t)\;dt\right)\overset{\textbf{FTC}}{=}e^x-f(x).$

Solving

$\begin{cases} f''(x)=e^x-f(x) \\f(0)=1\\f'(0)=1 \end{cases}\quad\quad\quad(\star\star)$

gives

$f(x) = \frac{1}{2}(\sin(x)+\cos(x)+e^x).$

Fig. 1

Notice the derivation of $(\star\star)$ can be simplified if Leibniz’s Rule (LR-1, see “A Semi-Rigorous Derivation of Leibniz’s Rule”) is applied:

$\frac{df(x)}{dx} = e^x + \underline{\frac{d}{dx}\int\limits_{0}^{x}(t-x)f(t)\;dt}$

$\overset{\textbf{LR-1}}{=} e^x + \underline{\int\limits_{0}^{x}\frac{\partial}{\partial x}(t-x)f(t)\;dt}$

$=e^x+\int\limits_{0}^{x}-1\cdot f(t)\;dt$

$= e^x-\int\limits_{0}^{x}f(t)\;dt$

$\implies \frac{d^2f(x)}{dx}=e^x-\frac{d}{dx}\int\limits_{0}^{x}f(t)\;dt\overset{\textbf{FTC}}{=}e^x-f(x).$

Fig.2 shows that Omega CAS explorer‘s Maxima engine is both FTC and LR-1 aware:

Fig. 2

Exercise-1 Given:

$f(x) = \int\limits_{0}^{x}t\cdot f(x-t)\;dt+\sin(x)$

where $f(x)$ is a continuous function, find $f(x).$

hint: Let $u=x-t, t = x-u; t=0\implies u=x; t=x\implies u=0; \frac{du}{dt}=-1$;

$f(x) = \int\limits_{x}^{0}(x-u)\cdot f(u)\cdot (-1)\;du + \sin(x)=\int\limits_{0}^{x}(x-u)f(u)\;du+\sin(x).$

# FTC saves the day!

Problem-1 Given

$f(x) = \int\limits_{0}^{2x}f(\frac{t}{2})\;dt +\log(2)\quad\quad\quad(\star)$

where $f(x)$ is a continuous function, find $f(x).$

Solution

Let

$p=2x \implies \frac{dp}{dx} = 2.\quad\quad\quad(1-1)$

By $(\star)$,

$\frac{df(x)}{dx} =\frac{d}{dx} \int\limits_{0}^{p} f(\frac{t}{2})\;dt + \frac{d\log(2)}{dx} = \underline{\frac{d}{dp}\left(\int\limits_{0}^{p} f(\frac{t}{2})\;dt\right)} \cdot \frac{dp}{dx}\overset{\textbf{FTC}}{=}\underline{f(\frac{p}{2})}\cdot \frac{dp}{dx}\overset{(1-1)}{=}2f(x),$

i.e.,

$\frac{df(x)}{dx} = 2f(x).\quad\quad\quad(1-2)$

Moreover, we see from $(\star)$ that

$f(0) = \int\limits_{0}^{0}f(\frac{t}{2})\;dt + \log(2) = 0 + \log(2) = \log(2).\quad\quad\quad(1-3)$

Solving initial-value problem

$\begin{cases} \frac{df(x)}{dx} = 2f(x)\\ f(0)=\log(2)\end{cases}$

gives

$f(x) = \log(2)\cdot e^{2x}.$

We use Omega CAS Explorer to verify:

Fig. 1-1

Problem-2 Given

$\int\limits_{0}^{1}f(u\cdot x) \;du = \frac{1}{2} f(x) +1\quad\quad\quad(\star\star)$

where $f(x)$ is a continuous function, find $f(x).$

Solution

Let $p=u\cdot x,$

$u=\frac{p}{x} \implies \frac{du}{dp} = \frac{1}{x}\quad\quad\quad(2-1)$

$u=0\implies p=0; u=1\implies p=x.\quad\quad\quad(2-2)$

$\int\limits_{0}^{1}f(u\cdot x)\;du\overset{(2)}{=} \int\limits_{0}^{x}f(p)\cdot\frac{du}{dp}\cdot dp\overset{(1)}{=}\int\limits_{0}^{x}f(p)\frac{1}{x}\;dp=\frac{1}{x}\int\limits_{0}^{x}f(p)\;dp.\quad\quad\quad(2-3)$

By (2-3), we express $(\star\star)$ as

$\frac{1}{x}\int\limits_{0}^{x}f(p)\;dp = \frac{1}{2}f(x)+1,$

i.e.,

$\int\limits_{0}^{x} f(p)\;dp = \frac{x}{2}f(x)+x.$

It follows that

$\underline{\frac{d}{dx}\left(\int\limits_{0}^{x}f(p)\;dp\right)}=\frac{d}{dx}\left(\frac{x}{2}f(x)+x\right)\overset{\textbf{FTC}}{\implies}\underline{f(x)}=\frac{1}{2}\left(f(x) + x\frac{d f(x)}{dx}\right)+1.\;(2-4)$

Solving differential equation (2-4) (see Fig. 2-1) gives

$f(x) = c x + 2.$

Fig. 2-1

The solution is verified by Omega CAS Explorer:

Fig. 2-2

Exercise-1 Solving $\begin{cases} \frac{df(x)}{dx} = 2f(x)\\ f(0)=\log(2)\end{cases}$ using a CAS.

Exercise-2 Solving (2-4) without using a CAS.

# An Epilogue to “Truth vs. Intellect”

This post illustrates an alternative of compute the approximate value of $\pi$.

We begin with a circle whose radius is $r$, and let $L_{n}, L_{n+1}$ denotes the side’s length of regular polygon inscribed in the circle with $2^n$ and $2^{n+1}$ sides respectively, $n=2, 4, ....$

Fig. 1

On one hand, we see the area of $\Delta ABC$ as

$\frac{1}{2}\cdot AB\cdot BC = \frac{1}{2}\cdot AB\cdot L_{n+1}$.

On the other hand, it is also

$\frac{1}{2}\cdot AC\cdot BE = \frac{1}{2}\cdot 2r\cdot \frac{L_n}{2}=\frac{1}{2}\cdot r\cdot L_n.$

Therefore,

$\frac{1}{2}AB\cdot L_{n+1}= \frac{1}{2}r\cdot L_n.$

Or,

$AB^2\cdot L_{n+1}^2 = r^2\cdot L_n^2\quad\quad\quad(1)$

where by Pythagorean theorem,

$AB^2= (2r)^2 - L_{n+1}^2.\quad\quad\quad(2)$

Substituting (2) into (1) gives

$(4r^2-L_{n+1}^2)L_{n+1}^2 = L_n^2\implies 4r^2L_{n+1}^2 - L_{n+1}^4 = r^2 L_n^2.$

That is,

$L_{n+1}^4-4r^2L_{n+1}^2+r^2 L_n^2 = 0.$

Let $p = L_{n+1}^2$, we have

$p^2-4r^2 p + r^2 L_n^2=0.\quad\quad\quad(3)$

Solving (3) for $p$ yields

$p = 2r^2 \pm r \sqrt{4 r^2-L_n^2}.$

Since $L_n^2$ must be greater than $L_{n+1}^2$ (see Exercise 1), it must be true (see Exercise 2) that

$L_{n+1}^2=2r^2 - r \sqrt{4r^2-L_n^2}.\quad\quad\quad(4)$

Notice when $r=\frac{1}{2}$, we obtain (5) in “Truth vs. Intellect“.

With increasing $n$,

$L_n\cdot 2^n \approx \pi\cdot 2r \implies \pi \approx \frac{L_n 2^n}{2r}.\quad\quad\quad$

We can now compute the approximate value of $\pi$ from any circle with radius $r$:

Fig. 2 $r=2$

Fig. 3 $r=\frac{1}{8}$

Exercise 1 Explain $L_{n}^2 > L_{n+1}^2$ geometrically.

Exercise 2 Show it is $2r^2-r\sqrt{4r^2-L_n^2}$ that represents $L_{n+1}^2.$

# Truth vs. Intellect

It was known long ago that $\pi$, the ratio of the circumference to the diameter of a circle, is a constant. Nearly all people of the ancient world used number $3$ for $\pi$. As an approximation obtained through physical measurements with limited accuracy, it is sufficient for everyday needs.

An ancient Chinese text (周髀算经,100 BC) stated that for a circle with unit diameter, the ratio is $3$.

In the Bible, we find the following description of a large vessel in the courtyard of King Solomon’s temple:

He made the Sea of cast metal, circular in shape, measuring ten cubits from rim to rim and five cubits high, It took a line of thirty cubits to measure around it. (1 Kings 7:23, New International Version)

This infers a value of $\pi = \frac{30}{10} = 3$.

It is fairly obvious that a regular polygon with many sides is approximately a circle. Its perimeter is approximately the circumference of the circle. The more sides the polygon has, the more accurate the approximation.

To find an accurate approximation for $\pi$, we inscribe regular polygons in a circle of diameter $1$. Let $L_{n}, L_{n+1}$ denotes the side’s length of regular polygon with $2^n$ and $2^{n+1}$ sides respectively, $n=2, 4, ....$

Fig. 1

From Fig. 1, we have

$\begin{cases} L_{n+1}^2 = x^2 + (\frac{1}{2} L_n)^2\quad\quad\quad(1) \\ (\frac{1}{2})^2 = (\frac{1}{2}L_n)^2 + y^2\quad\quad\;\quad(2)\\ x+y = \frac{1}{2}\;\quad\quad\quad\quad\quad\quad(3) \end{cases}$

It follows that

$y\overset{(2)}{=}\sqrt{(\frac{1}{2})^2-(\frac{1}{2} L_n)^2} \overset{(3)}{ \implies} x=\frac{1}{2}-\sqrt{(\frac{1}{2})^2-(\frac{1}{2}L_n)^2}.\quad\quad\quad(4)$

Substituting (4) into (1) yields

$L_{n+1}^2 = \left(\frac{1}{2}-\sqrt{(\frac{1}{2})^2-(\frac{1}{2}L_n)^2}\right)^2+(\frac{1}{2}L_n)^2$

That is,

$L_{n+1}^2 = \frac{1}{4}\left(L_n^2 + \left(1-\sqrt{1-L_n^2}\right)^2\right).$

Further simplification gives

$L_{n+1}^2 = \frac{1}{2}\left(1-\sqrt{1-L_n^2}\right),\quad\quad\quad(5)$

Starting with an inscribed square $(L_2^2 =\frac{1}{2})$, we compute $L_{n+1}^2$ from $L_{n}^2$ (see Fig. 2). The perimeter of the polygon with $2^{n+1}$ sides is $2^{n+1} \cdot L_{n+1}$.

Fig. 2

Clearly,

$\lim\limits_{n \rightarrow \infty} 2^n \cdot L_{n} = \pi$.

Exercise-1 Explain, and then make the appropriate changes:

Hint: (5) is equivalent to $L_{n+1}^2 = \frac{L_n^2}{2\left(1+\sqrt{1-L_{n}^2}\right)}.$

# We all bleed the same color

In “Mathematical Models in Biology”, Leah Edelstein-Keshet presents a model describing the number of circulating red blood cells (RBC’s). It assumes that the spleen filters out and destroys a fraction of the cells daily while the bone marrow produces a amount proportional to the number lost on the previous day:

$\begin{cases} R_{n+1} = (1-f)R_n+M_n\\ M_{n+1} = \gamma f R_n\end{cases}(1)$

where

$R_n -$ number of RBC’s in circulation on day $n$,

$M_n -$ number of RBC’s produced by marrow on day $n$,

$f -$ fraction of RBC’s removed by the spleen,

$\gamma -$ numer of RBC’s produced per number lost.

What would be the cell count on the $n^{th}$ day?

Observe first that (1) is equivalent to

$R_{n+2} = (1-f)R_{n+1}+M_{n+1}\quad\quad\quad(2)$

where

$M_{n+1} = \gamma f R_n.\quad\quad\quad(3)$

Let $n = -1$,

$M_0=\gamma f R_{-1} \implies R_{-1} = \frac{M_0}{\gamma f}.\quad\quad\quad(4)$

Substituting (3) into (2) yields

$R_{n+2} = (1-f)R_{n+1}+\gamma f R_{n}.$

We proceed to solve the following initial-value problem using ‘solve_rec‘ (see “Solving Difference Equations using Omega CAS Explorer“):

$\begin{cases} R_{n+2}=(1-f)R_{n+1}+\gamma f R_{n}\\ R_{0}=1, R_{-1} = \frac{1}{\gamma f}\end{cases}$

Evaluate the solution with $f=\frac{1}{2}, g=1$, we have

$R_n = \frac{4}{3} + \frac{(-1)^{n+1}2^{-n}}{3}.\quad\quad\quad(5)$

Plotting (5) by ‘plot2d(4/3 + (-1)^(n+1)*2^(-n)/3, [n, 0, 10], WEB_PLOT)’ fails (see Fig. 1) since plot2d treats (5) as a continuous function whose domain includes number such as $\frac{1}{2}$.

Fig. 1

Instead, a discrete plot is needed:

Fig. 2

From Fig. 2 we see that $R_{n}$ converges to a value between $1.3$ and $1.35$. In fact,

$\lim\limits_{n \rightarrow \infty} \frac{4}{3} + \frac{(-1)^{n+1}2^{-n}}{3} = \frac{4}{3}\approx 1.3333....$

# Solving Difference Equations using Omega CAS Explorer

Maxima’s ‘solve_rec‘ solves difference equation:

Fig. 1 Geometric Sequence

It solves initial-value problem as well:

Fig. 2 Fibonacci Sequence

Mathematica has ‘RSolve‘:

and ‘RSolveValue‘:

# Fire & Water

Question:

When a forrest fire occurs, how many fire fighter should be sent to fight the fire so that the total cost is kept minimum?

Suppose the fire starts at $t=0$; at $t=t_1$, the fire fighter arrives; the fire is extinguished later at $t=t_2$.

Let $c_1, c_2, c_3$ and $x$ denotes the monetary damage per square footage burnt, the hourly wage per fire fighter, the one time cost per fire fighter and the number of fire fighters sent respectively. The total cost consists damage caused by the fire, the wage paid to the fire fighters and the one time cost of the fire fighters:

$c(x) = c_1\cdot$(total square footage burnt) $+\;c_2\cdot (t_2-t_1)\cdot x + c_3\cdot x.$

Notice $t_2-t_1$, the duration of fire fighting.

Assume the fire ignites at a single point and quickly spreads in all directions with flame velocity $v_*$, the growing square footage of engulfed circular area is a function of time. Namely,

$b(t) = \pi r^2 = \pi (v_* t) ^2 = \pi v_*^2 t^2 \overset{k=\pi v_*^2}{=} k t^2.$

Its burning rate

$v_b(t) = b'(t) = 2 k t \overset{\alpha = 2 k}{=} \alpha t.\quad\quad\quad(1)$

However, after the arrival of $x$ fire fighters, $\alpha$ is reduced by $\beta x$, an amount that is directly proportional to the number of fire fighters on the scene. The reduction of $\alpha$ reflects the fire fighting efforts exerted by the crew. As a result, for $t > t_1, v_b(t)$ declines along the line described by

$v_b(t) = (\alpha-\beta x)t + d$

where $\alpha-\beta x <0$. Or equivalently,

$\beta x - \alpha >0.\quad\quad\quad(2)$

Moreover, the fire is extinguished at $t_2$ suggests that

$v_b(t_2) = 0 \implies (\alpha-\beta x) t_2 + d =0 \implies d = -(\alpha-\beta x)t_2.$

i.e.,

$\forall t_1< t \le t_2, v_b(t) = (\alpha-\beta x) t - (\alpha-\beta x) t_2=(\alpha-\beta x) (t-t_2).\quad\quad\quad(3)$

Combine (1) and (3),

$v_b(t) = \begin{cases} \alpha t, \quad\quad\quad\quad\quad\quad\quad\quad0\le t \le t_1\\ (\alpha-\beta x)(t-t_2),\quad\quad t_1< t \le t_2 \end {cases}.$

It is further assumed that

$v_b(t)$ is continuous at $t=t_1.\quad\quad\quad(4)$

We illustrate $v_b(t)$ in Fig. 1.

Fig. 1

The fact that $v_b(t)$ is continuous at $t_1$ means

$\lim\limits_{t \to t_1^+}v_b(t)=(\alpha-\beta x)(t_1-t_2) = \lim\limits_{t \to t_1^-}v_b(t) = h \implies (\alpha-\beta x) (t_1-t_2)=h.$

That is,

$t_2-t_1=\frac{h}{\beta x - \alpha}.\quad\quad\quad(5)$

The area of triangle in Fig. 1 represents $b(t_2)$, the total square footage damaged by the fire. i.e.,

$b(t_2) =\frac{1}{2}t_1 h + \frac{1}{2}(t_2-t_1)h\overset{(5)}{=}\frac{1}{2}t_1 h+ \frac{1}{2}\frac{h^2}{\beta x-\alpha}.\quad\quad\quad(6)$

Consequently, the total cost

$c(x)= c_1 b(t_2) + c_2 x (t_2-t_1) + c_3 x \overset{(6), (5)}{=}\frac{c_1 t_1 h}{2} + \frac{c_1 h^2}{2(\beta x-\alpha)} + \frac{c_2 h x}{\beta x-\alpha} + c_3 x.$

To minimize the total cost, we seek a value $x$ where function $c$ attains its minimum value $c(x)$.

From expressing $x$ as

$x = \frac{1}{\beta}\beta x = \frac{1}{\beta}({\beta x-\alpha +\alpha} )= \frac{1}{\beta}(\beta x -\alpha) + \frac{\alpha}{\beta},$

we obtain

$\frac{c_2 h x}{\beta x-\alpha} = \frac{c_2 h}{\beta x-\alpha}(\frac{1}{\beta}(\beta x - \alpha) + \frac{\alpha}{\beta})=\frac{c_2 h}{\beta} + \frac{c_2 \alpha h}{\beta(\beta x-\alpha)},\quad\quad\quad(7)$

$c_3 x = \frac{c_3}{\beta}(\beta x-\alpha) + \frac{c_3 \alpha}{\beta}.\quad\quad\quad(8)$

It follows that

$c(x) = \underbrace{\frac{c_1 t_1 h}{2} + \frac{c_1 h^2}{2(\beta x-\alpha)}}_{c_1\cdot(6)} + \underbrace{\frac{c_2 h}{\beta}+\frac{c_2 \alpha h}{\beta(\beta x - \alpha)}}_{(7)} + \underbrace{\frac{c_3}{\beta}(\beta x-\alpha) + \frac{c_3 \alpha}{\beta}}_{(8)}$

$=\frac{c_1 t_1 h}{2} + \frac{c_2 h}{\beta} + \frac{c_3 \alpha}{\beta} + \underline{\frac{c_1 \beta h^2+2 c_2 \alpha h}{2\beta(\beta x - \alpha)} + \frac{c_3(\beta x-\alpha)}{\beta}}.$

Hence, to minimize $c(x)$ is to find the $x$ that minimizes $\frac{c_1\beta h^2+2 c_2\alpha h }{2\beta(\beta x - \alpha)} + \frac{c_3(\beta x - \alpha)}{\beta}.$

Since $\frac{c_1\beta h^2+2 c_2\alpha h }{2\beta(\beta x - \alpha)} \cdot \frac{c_3(\beta x - \alpha)}{\beta}=\frac{(c_1 \beta h^2+2 c_2 \alpha h)c_3}{2\beta^2}$, a constant, by the following theorem:

For positive quantities $a_1, a_2, ..., a_n, c_1, c_2, ..., c_n$ and positive rational quantities $p_1, p_2, ..., p_n$, if $a_1^{p_1}a_2^{p_2}...a_k^{p_k}$ is a constant, then $c_1a_1+c_2a_2+...+c_na_n$ attains its minimum if $\frac{c_1a_1}{p_1} = \frac{c_2a_2}{p_2} = ... = \frac{c_na_n}{p_n}.$

(see “Solving Kepler’s Wine Barrel Problem without Calculus“), we solve equation

$\frac{c_1\beta h^2+2 c_2\alpha h }{2\beta(\beta x - \alpha)} = \frac{c_3(\beta x - \alpha)}{\beta}$

for $x$:

Fig. 2

From Fig. 2, we see that $x =\frac{\alpha}{\beta} - \frac{\sqrt{\beta c_1 c_3 h^2+2\alpha c_2 c_3 h}}{\sqrt{2}\beta c_3} \implies \beta x - \alpha < 0$, contradicts (2).

However, when $x =\frac{\sqrt{\beta c_1 c_3 h^2+2\alpha c_2 c_3 h}}{\sqrt{2}\beta c_3} + \frac{\alpha}{\beta}=\sqrt{\frac{\beta c_1 h^2 + 2 \alpha c_2 h}{2\beta^2 c_3}} + \frac{\alpha}{\beta}, \beta x-\alpha$ is a positive quantity. Therefore, it is

$x \overset{h=\alpha t_1}{=} \alpha \sqrt{\frac{\beta c_1 t_1^2 + 2 c_2 t_1}{2\beta^2 c_3}} + \frac{\alpha}{\beta}$

that minimizes the total cost.

# A Not So Sinful Delight

Problem:

Show that if $x+\frac{1}{x}=1$ then $x^7+\frac{1}{x^7}=1.$

Solution-1:

From

$x+\frac{1}{x} = 1,\quad\quad\quad(0)$

we have

$x^2+1=x\quad\quad\quad(1)$

$\implies x^2=x-1\quad\quad\quad(2)$

$\implies x^3=x^2-x\overset{(2)}{=}(x-1)-x=-1\quad\quad\quad(3)$

$\implies x^6=1\quad\quad\quad(4)$

$\implies x^7=x\quad\quad\quad(5)$

$\implies x^7+\frac{1}{x^7}\overset{(5)}{=}x+\frac{1}{x}\overset{(0)}{=}1.$

Solution-2:

Exercise-1 Given $x^3+4x=8$, determine the value of $x^7+64x^2.$

# Algebra, CAS vs Human

Problem:

Given $x^3+4x=8$, determine the value of $x^7+64x^2$.

Solution-1 (CAS)

Solution-2 (Human)

From

$x^3+4x=8,\quad\quad\quad(1)$

we have

$(x^3+4x)^2=8^2$

$\implies x^6+8x^4+16x^2=64$

$\implies x^7+8x^5+16x^3=64x$

$\implies x^7+8x^5+16x^3+16x^3=64x +16x^3$

$\implies x^7+8x^5+32x^3=64x+16x^3$

$\implies x^7+8x^2(x^3+4x) =16(x^3+4x)$

$\overset{(1)}{\implies} x^7+8x^2\cdot 8 = 16\cdot 8$

$\implies x^7+64x^2=128$

Exercise-1 Show that if $x+\frac{1}{x} =1$ then $x^7+\frac{1}{x^7}=1.$