When Least is Best

July 26, 2023August 3, 2023 / Michael Xue / Leave a comment

A business sells a product has the following process:

Product items are stocked at the beginning of a period. $c_1$ is the stocking fee ($). The items are then sold at the rate $r$ (# of items per unit time). When all items are sold, the period ends.

The process repeats indefinitely.

Fig. 1

Let $q(t)$ denotes the inventory level at time $t$ . Then

$q(t) = q_0 - rt$

where $q_0$ is the number of product stocked at the beginning.

Let $T$ be the duration of the period. We have

$q(T) = q_0-rT=0 \implies q_0= rT\quad\quad\quad(1)$

since at $T$ , all products are sold.

Fig. 2

The cost for having an inventory during the period $T$ is

$\displaystyle\int\limits_{0}^{T} q(t)\cdot c_2 \;dt = \int\limits_{0}^{T}(q_0-rt)c_2 dt = c_2\int\limits_{0}^{T}(q_0-rt)\;dt$

where $c_2$ ($ per item per unit time) pays for inventory space and maintenance.

It is not necessary to evaluate the definite integral for it is simply the area of the shaded triangle in Fig. 3

Fig. 3

which is

$\frac{1}{2}q_0T.$

Therefore, the total cost for the period

$c= c_1 + c_2\cdot\frac{1}{2}q_0T \overset{(1)}{=} c_1+c_2\cdot\frac{1}{2}(rT)\cdot T=c_1 + \frac{1}{2}c_2rT^2.$

And, the total cost per unit time for the period

$\bar{c} = \frac{c}{T} = \frac{c_1}{T} + \frac{1}{2}c_2rT.\quad\quad\quad(2)$

Our objective is to determine $q_0$ so that $\bar{c}$ is minimized.

From (2), we see $\frac{c_1}{T}\cdot\frac{1}{2}c_2rT = \frac{1}{2}c_1c_2r,$ a constant. By Theorem-2 in “Solving Kepler’s ‘Wine Barrel Problem’ without Calculus“, when

$\frac{c_1}{T} = \frac{1}{2}c_2rT,\quad\quad\quad(3)$

$\bar{c}$ attains its minimum.

Solving (3) for $T$ gives

$T^* = \sqrt{\frac{2c_1}{c_2r}}.$

It follows that by (1),

$q_0^* = r\sqrt{\frac{2c_1}{c_2r}} = \sqrt{\frac{2c_1r}{c_2}}.$

We can obtain $T^*, q_0^*$ and $\bar{c}^*$ by Calculus:

$\frac{d\bar{c}}{dT} = (\frac{c_1}{T} \frac{1}{2}c_2rT)' = -\frac{c_1}{T^2} + \frac{1}{2}c_2r$

$\frac{d\bar{c}}{dT} = 0 \implies T = \sqrt{\frac{2c_1}{c_2r}}$

$\frac{d^2\bar{c}}{dT^2} = \frac{d}{dT}(\frac{d\bar{c}}{dT}) = (-c_1T^{-2})' = 2c_1T^{-3} > 0\;\forall T>0.$

$T^*=\sqrt{\frac{2c_1}{c_2r}},$

$q_0^* = r T^* = r\sqrt{\frac{2c_1}{c_2r}} = \sqrt{\frac{2c_1r}{c_2}},$

$\bar{c}^* = \sqrt{2c_1c_2r}.$

We can also determine the values free of derivative from solving an inequality:

$\bar{c} = \frac{c_1}{T}+\frac{1}{2}c_2rT$

$\implies \bar{c}T = c_1+\frac{1}{2}c_2rT^2$

$\implies \frac{1}{2}c_2rT^2-\bar{c}T+c_1 =0$

$\implies \Delta = \bar{c}^2-4\cdot\frac{1}{2}c_2r\cdot c_1 = \bar{c}^2-2c_2rc_1\ge 0$

$\implies \bar{c} \ge \sqrt{2c_1c_2r}$

$\implies \bar{c}^* = \sqrt{2c_1c_2r}.$

It follows that

$T^* = \frac{-(-\bar{c}^*)}{2\cdot\frac{1}{2}c_2\cdot r}=\frac{\sqrt{2c_1c_2r}}{c_2r} = \sqrt{\frac{2c_1}{c_2r}},$

$q_0^* = rT^* = \sqrt{\frac{2c_1r}{c_2}}.$

Prelude to Construct an Epsilon-Delta Style Proof

July 21, 2023July 24, 2023 / Michael Xue / Leave a comment

Deduce from the result obtained in “cos(x), sin(x)/x and 1 (Part 2)“, namely,

$\forall x \ni 0< |x| <\frac{\pi}{2}, \cos(x) <\frac{\sin(x)}{x} < 1\quad\quad\quad(*)$

the following:

$\forall x \ni 0<|x|< \frac{\pi}{2}, \underline{1-x^2}<\frac{\sin(x)}{x} < 1.$

Since

$1-\cos(x) = (1-\cos(x))\frac{1+\cos(x)}{1+\cos(x)}=\frac{\sin(x)^2}{1+\cos(x)}<\sin(x)^2 \overset{(*)}{<} x^2,$

we have

$1-\cos(x) < x^2 \implies \underline{1-x^2< \cos(x)}.$

Rewrite (*) as

$0<|x| < \frac{\pi}{2}, \underline{1-x^2<\cos(x)}<\frac{\sin(x)}{x}<1$

gives

$\forall x \ni 0<|x| <\frac{\pi}{2}, 1-x^2 < \frac{\sin(x)}{x} < 1.$

From Fig. 1 in “cos(x), sin(x)/x and 1 (Part 1)“:

we see

$L=x\implies L^2 = x^2 > \underbrace{BC^2=1^2+1^2-2\cdot 1\cdot 1\cdot \cos(x)}_\mathrm{cosine\; law}=2+2\cos(x).$

That is,

$x^2 > 2+2\cos(x)$

which yields

$1-\frac{x^2}{2} < \cos(x) \overset{1-x^2< 1-\frac{x^2}{2}}{\implies} 1-x^2<\cos(x).$

Therefore, by (*),

$\forall x \ni 0<x<\frac{\pi}{2}, 1-x^2<\frac{\sin(x)}{x}<1.\quad\quad\quad(1)$

Morever, $\frac{\pi}{2} < x < 0$ means $0<\boxed{-x}<\frac{\pi}{2}.$ By (*),

$1-(\boxed{-x})^2 < \frac{\sin(\boxed{-x})}{\boxed{-x}} < 1.$

$(\boxed{-x})^2=x^2, \frac{\sin(\boxed{-x})}{\boxed{-x}} = -\frac{\sin(x)}{-x}=\frac{\sin(x)}{x}$ then gives

$\forall x \ni -\frac{\pi}{2} < x <0, 1-x^2 < \frac{\sin(x)}{x}<1.\quad\quad\quad(2)$

Putting (1) and (2) together,

$\forall x \ni |x|<\frac{\pi}{2}, 1-x^2 <\frac{\sin(x)}{x} < 1 .$

When x->0, sin(x)/x->1

July 20, 2023July 20, 2023 / Michael Xue / Leave a comment

We have obtained the following previously:

$\forall x \ni 0 < |x| < \frac{\pi}{2},\cos(x) <\frac{\sin(x)}{x}<1.$

(see “cos(x), sin(x)/x and 1 (Part 2)“)

That is,

$\exists \delta = \frac{\pi}{2} \ni 0 < |x-0| < \delta, \underbrace{f(x)}_{\cos(x)} < \underbrace{g(x)}_{\frac{\sin(x)}{x}} <\underbrace{h(x)}_{1}.$

Since $\lim\limits_{x \rightarrow 0} f(x) = \lim\limits_{x \rightarrow 0} \cos(x)= 1, \lim\limits_{x \rightarrow 0} h(x) = \lim\limits_{x \rightarrow 0}1 = 1,$ we have

$\begin{cases}\exists \delta >0 \ni 0 < |x-0|<\delta, f(x) < g(x) <h(x), \\ \lim\limits_{x \rightarrow 0} f(x) =\lim\limits_{x \rightarrow 0} h(x) = L=1\end{cases}$

Therefore, by “Sandwich Theorem for Functions“,

$\lim\limits_{x \rightarrow 0} g(x) = \lim\limits_{x \rightarrow 0}\frac{\sin(x)}{x} = 1.$

cos(x), sin(x)/x and 1 (Part 2)

July 19, 2023July 19, 2023 / Michael Xue / Leave a comment

$\forall x \ni 0 < |x| < \frac{\pi}{2}, \cos(x) < \frac{\sin(x)}{x} < 1.\quad\quad\quad(*)$

What (*) states is that

$0< x < \frac{\pi}{2}, \cos(x) <\frac{ \sin(x)}{x} < 1\quad\quad\quad(1)$

and

$-\frac{\pi}{2} < x < 0, \cos(x) < \frac{\sin(x)}{x} < 1.\quad\quad\quad(2)$

We have shown that (1) is true in “cos(x), sin(x)/x and 1 (Part 1)“.

Let’s prove (2) now:

$-\frac{\pi}{2} < x < 0$ means $0 <\boxed{ -x} < \frac{\pi}{2}.$ So by (1),

$0 < \boxed{-x} < \frac{\pi}{2}, \cos(\boxed{-x}) < \frac{\sin(\boxed{-x})}{\boxed{-x}} < 1.\quad\quad\quad(3)$

Since $\cos(\boxed{-x}) = \cos(x), \frac{\sin(\boxed{-x})}{\boxed{-x}} = \frac{-\sin(x)}{-x} = \frac{\sin(x)}{x},$ (3) gives

$-\frac{\pi}{2}< x < 0, \cos(x) < \frac{\sin(x)}{x} < 1.\quad\quad\quad(4)$

Putting (1) and (4) together, we have

$0 < |x| <\frac{\pi}{2}, \cos(x) < \frac{\sin(x)}{x} < 1.$

cos(x), sin(x)/x and 1 (Part 1)

July 18, 2023July 19, 2023 / Michael Xue / Leave a comment

$\forall x \in (0, \frac{\pi}{2}), \; \cos(x) <\frac{\sin(x)}{x} < 1.$

Fig. 1 $x=\frac{L}{r} = \frac{L}{1} = L \implies L = x$

From Fig. 1, we see

$A_{\Delta OCB} < A_{\overset{\large\frown}{OCB}} < A_{\Delta OAB}.$

i.e.,

$\frac{1}{2}\cdot 1\cdot \sin(x) < \frac{1}{2}\cdot 1^2\cdot x < \frac{1}{2}\cdot 1\cdot \tan(x).$

That is,

$\sin(x) <x < \tan(x).$

Since $\sin(x) < x, x>0,$

$\frac{\sin(x)}{x}<1.\quad\quad\quad(1)$

And, for $x \in (0, \frac{\pi}{2}), \cos(x) > 0.$ Therefore, $x < \tan(x)=\frac{\sin(x)}{\cos(x)}$ gives

$\cos(x) < \frac{\sin(x)}{x}.\quad\quad\quad(2)$

Putting (1) and (2) together, we have

$\cos(x) < \frac{\sin(x)}{x} < 1.$

Sandwich Theorem for Functions

July 1, 2023July 4, 2023 / Michael Xue / Leave a comment

Prove:

$\begin{cases} \exists \delta \ni 0 < |x-a| <\delta, f(x) \le g(x) \le h(x)\quad\quad(1) \\ \lim\limits_{x \rightarrow a}f(x) = \lim\limits_{x \rightarrow a}h(x) = L \quad\quad\quad\quad\quad\quad\quad\quad\quad(2)\end {cases} \implies \lim\limits_{x \rightarrow a} g(x) = L.$

We express (1) as

$\exists \delta \ni 0 <|x-a|<\delta \implies f(x) \le g(x) \le h(x)$

and (2):

$\forall \epsilon > 0, \exists \delta_1 \ni 0<|x-a|<\delta_1 \implies |f(x)-L| < \epsilon$

$\forall \epsilon > 0, \exists \delta_2 \ni 0<|x-a|<\delta_2 \implies |h(x)-L| < \epsilon$

Let $\delta_* = \min(\delta, \delta_1, \delta_2),$

we have

$\exists \delta_* \ni 0 <|x-a|<\delta_* \implies f(x) \le g(x) \le h(x).$

$\forall \epsilon > 0, \exists \delta_* \ni 0<|x-a|<\delta_* \implies |f(x)-L| < \epsilon$

$\forall \epsilon > 0, \exists \delta_* \ni 0<|x-a|<\delta_* \implies |h(x)-L| < \epsilon$

It means

$\exists \delta_* \ni 0 <|x-a|<\delta_* \implies \underline{f(x) \le g(x) \le h(x)}$

$\forall \epsilon > 0, \exists \delta_* \ni 0<|x-a|<\delta_* \implies \underline{-\epsilon+L < f(x)} < \epsilon + L$

$\forall \epsilon > 0, \exists \delta_* \ni 0<|x-a|<\delta_* \implies -\epsilon +L < \underline{h(x) < \epsilon+L}$

Hence,

$\exists \delta_* \ni 0 <|x-a|<\delta_* \implies -\epsilon +L < f(x) \le g(x) \le h(x) < \epsilon + L$

That is

$\exists \delta_* \ni 0 <|x-a|<\delta_* \implies -\epsilon +L < g(x) < \epsilon + L$

i.e.,

$\exists \delta_*\ni 0 <|x-a|<\delta_* \implies |g(x) - L| < \epsilon$

It follows that

$\lim\limits_{x \rightarrow a } g(x) = L.$

See also “Sandwich Theorems and Their Proofs“.