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Elasticity refers to the percent change to which one value changes, quantity of $$x$$, when another does. Supply and demand change with respect to price; investment and savings change with respect to interest rate. The name is "X elasticity of Y" where a change in X causes a change of magnitude (the elasticity * Y).

Price elasticity of Demand
Let's begin with some definitions. If we want to determine the percent change in price for the demand for a good, we need to understand what we are doing. Let $$Q_{D0}$$ denote the quantity demanded at some arbitrary initial point, and let $$Q_{D}$$ denote the quantity demand at some point close to our initial point that is the final value we get to. To determine the percent change, we have to find a change in the values that we have chosen divided by our old value: if you want to put it in terms of what we learned, it is
 * $$%dQ_{D}={Q_{D} - Q_{D0} \over Q_{D0}}$$, where $$d$$ is simply a letter that denotes a very small change in $$Q$$ (quantity demand).

We learn the above information for one very special reason. That $$%d$$ means something: percentage change. We are looking at the percentage change in quantity demanded. However, we cannot find the percentage in quantity demanded without knowing the prices of goods that correspond to those values. Therefore, let $$P_{D}$$ be the price of $$Q_{D}$$ and $$P_{D0}$$ be the price of $$Q_{D0}$$. To determine the percent change, we have to find a change in the values that we have chosen divided by our old value: if you want to put it in terms of what we learned, it is


 * $$%d P = {P - P_0 \over P_0}$$, where $$P$$ is for price corresponding to $$Q$$.

For any ordered pair $$(Q_{D0},\,P_{D0})$$ and $$(Q_{D},\, P_{D})$$ that corresponds to a demand curve, the price elasticity of demand is $$

Before talking about applications, it is important to get an intuitive understanding of elasticity. Please note the use of elasticity. It is not an arbitrary term. Rather, it brings to mind the rubber band. The maximum stretching distance of the rubber band $$\Delta x$$ can be assigned a value $$a$$ or $$b$$, where $$a>b$$. As a collective, we would say that the rubber band with $$\Delta x=a$$ is relatively more elastic than $$\Delta x=b$$. Similarly, the "stretchiness" of a demand curve represents this fundamental relationship in the form of a percentage change in both quantity demanded and price using the greek letter eta ($$\eta_{D}$$).

Because the demand curve comes in all types of curves (that is downward-sloping), the elasticity must be negative. This does not change anything fundamental to how we interpret the value, yet the absolute value of the price elasticity of demand is often used. Keep in mind that the change is not some constant either but a percentage. This means we are not finding the slope. Instead, we are looking at percentage change. As a result, some strange properties come true. One of them is that moving down along a linear demand curve implies a non-constant elasticity. In fact, the change in the elasticity decreases further and further. A proof for this concept is given below:


 * Given: $${\Delta Q \over \Delta P}={1 \over m}$$ is the reciprocal slope of a linear demand curve with slope $$m$$.
 * Prove: the elasticity $$\eta_{D}$$ of a linear demand curve must be decreasing as price decreases.

Knowing $${\Delta Q \over \Delta P}={1 \over m}$$ is the reciprocal slope, multiplying by $${P \over Q}$$, which is the point $$(Q,P)$$ that is variable, makes the expression equal to the elasticity of demand $$\eta_{D}$$. The reason for this is because

$$\begin{align} \frac{\Delta Q}{\Delta P}\cdot\frac{P}{Q} &= \frac{P\Delta Q}{Q\Delta P} \\ &= \frac{\Delta Q}{Q}\cdot\frac{P}{\Delta P} \\ &= \frac{\tfrac{\Delta Q}{Q}}{\tfrac{\Delta P}{P}} \end{align}$$.

That is,

$$

As you move down the demand curve, $$\Delta P<0$$ and $$\Delta Q>0$$. Ergo, $$\frac{\Delta P}{\Delta Q}<0$$. However, given $${\Delta Q \over \Delta P}$$ is a reciprocal slope of the linear demand curve, the value for $$\Delta Q$$ and $$\Delta P$$ are both constant. Hence, the slope $${1 \over m}$$ must be constant.

Summary: Because $${P\over Q}$$ decreases per every point $$P$$ to point $$Q$$, and $${1 \over m}$$ is constant, the elasticity of demand, $$\eta_{D}$$, along a demand curve must be falling as the price decreases.$$\blacksquare$$



Notice that we were able to derive one more equation relating to the elasticity of demand, $$\eta_{D}$$. However, we did not this proof to show them. By the very fact that$$\left\vert\eta_{D}\right\vert=\left\vert{%dQ_{D} \over %dP_{D}}\right\vert=\left\vert{{Q_{D} - Q_{D0} \over Q_{D0}} \over {P_{D}-P_{D0}\over P_{D0}}}\right\vert\!$$

we can further show one more new truth, simply through rearrangement. Here is the work that leads to the final equation:


 * $$\begin{align}

\left\vert\eta_{D}\right\vert&=\left\vert\frac{P_{D0}\left(Q_{D}-Q_{D0}\right)}{Q_{D0}\left(P_{D}-P_{D0}\right)}\right\vert\\ &=\frac{P_{D0}}{Q_{D0}}\cdot\left\vert\frac{Q_{D}-Q_{D0}}{P_{D}-P_{D0}}\right\vert\\ &=\frac{P_{D0}}{Q_{D0}}\cdot\left\vert\frac{dQ_{D}}{dP_{D}}\right\vert & \left[\text{Recall, }\frac{dQ_{D}}{dP_{D}}<0\text{.}\right] \end{align}$$

$$

Remember and  for later, especially equation 1.

Interpreting the Price Elasticity of Demand
We will look at the following demand curve above for this little exercise in price elasticity of demand. Keep note that the demand function, $$\operatorname{D_{i}}(Q)$$, is the solid, black, downward-sloping line. We will talk about the other two functions later.

Use of Price Elasticity of Demand
Just by knowing a numerical value, one can know more about the type of good that is possibly given in the market. However, before an honest analysis can be given for one particular good, the effects need to be given first before we can move on.

Price elasticity of Supply
By using the same logic for supply, the price elasticity of supply is $${{%\Delta S \over %\Delta P}={{S_{q,f} - S_{q,i} \over S_i} \over {P_f - P_i \over P_i}}}\!$$

where $$S_q$$ denotes the quantity supplied, $$S_{q,i}$$ denotes the quantity supplied at some arbitrary initial point along the demand curve, and $$S_{q,f}$$ denotes the quantity supplied at some final point close to $$S_{q,i}$$.

To reiterate why we determine price elasticity, the percent change can help us determine by how much our good has increased in quantity demanded or quantity supplied and can determine not only the slope but also the iterative change of a good. Say that an ordered pair exists in which $$(S_{q,i},\, P_i)$$ and $$(D_{q,f},\, P_f)$$ corresponds to a supply curve.

Note:
 * $$P_i$$ is initial price
 * $$P_f$$ is final price
 * $$S_{q,i}$$ is initial supply
 * $$S_{q,f}$$ is final supply
 * $$D_i$$ is initial demand
 * $$D_f$$ is final demand
 * $$r_i$$ is initial interest
 * $$r_f$$ is final interest
 * $$I_i$$ is initial investment
 * $$I_f$$ is final investment
 * $$S_i$$ is initial savings
 * $$S_f$$ is final savings

Price elasticity of demand


 * $$= \frac{ %change in D }{ %change in P } = \frac{ \frac{ D_f - D_i }{ ( D_f + D_i ) / 2 } }{ \frac{ P_f - P_i }{ ( P_f + P_i ) / 2 } } $$

Price elasticity of supply


 * $$= \frac{ %change in S }{ %change in P } = \frac{ \frac{ S_f - S_i }{ ( S_f + S_i ) / 2 } }{ \frac{ P_f - P_i }{ ( P_f + P_i ) / 2 } } $$

Interest elasticity of investment


 * $$= \frac{ %change in I }{ %change in r } = \frac{ \frac{ I_f - I_i }{ ( I_f + I_i ) / 2 } }{ \frac{ r_f - r_i }{ ( r_f + r_i ) / 2 } } $$

Interest elasticity of savings


 * $$= \frac{ %change in S }{ %change in r } = \frac{ \frac{ S_f - S_i }{ ( S_f + S_i ) / 2 } }{ \frac{ r_f - r_i }{ ( r_f + r_i ) / 2 } } $$

Cross elasticities
The elasticities mentioned above refer to one object. Cross elasticities refer to the effects of something's price, interest, etc. on something else. This comes into play with substitute and complementary goods and services for the consumer

Rational and Absolute Value Problem
{Find $$x$$: $$\displaystyle -\frac{x}{2\sqrt{4-x}}+\sqrt{4-x}=0$$ { 4 _5}
 * type="{}"}

{ Recall the Pythagorean Theorem: $$\displaystyle \sqrt{a^{2}+b^{2}}=c$$. ''Note: the below problems are very difficult. Congratulations if you obtain the correct answer.'' }

{A step is a unit movement left (negative), right (positive), down (negative), or up (positive) from the starting point $$\displaystyle O(0,0)$$. Let $$\displaystyle s$$ be the distance travelled while $$\displaystyle D$$ is the distance from $$\displaystyle O$$. Assume the object can only repeat $$\displaystyle R$$ times for only one chosen directional movement, and the object can only move four steps from $$\displaystyle O$$ (e.g. If $$\displaystyle R=1$$, then up, right, up, left is allowed. If $$\displaystyle R=1$$, then up, right, up, up is not allowed because up is repeated twice, and up, right, right, up is not allowed because up is repeated once and right is repeated once). The maximum deviation in distance travelled and displacement for each allowed $$\displaystyle R$$ repetition is indicated by $$\displaystyle T(R)$$. Use the following information to answer items 2 - 3.}

{What is the domain, $$\displaystyle A$$ of $$\displaystyle T(R)$$ that is the most practical to use in this situation? - $$\displaystyle A=\{R|R=0,1,2,3,4\}$$ + $$\displaystyle A=\{R|R=0,1,2,3\}$$ - $$\displaystyle A=\{R|R=1,2,3,4\}$$ - $$\displaystyle A=\{R|R=1,2,3\}$$
 * type="" coef="3"}

{Which of the following best indicates the graph $$T(R)$$? - $$\displaystyle A\times\mathbb{N}$$ - $$\displaystyle A\times\mathbb{Z}$$ - $$\displaystyle A\times\mathbb{Q}$$ + $$\displaystyle A\times\mathbb{R}$$ For Exploration 3-1 through Exploration 3-4, $$f(x)$$ is an absolute value function, and g(x) is any arbitrary function.
 * type="" coef="3"}

Logarithm and Summation Problem
{Find the sum of each of the following:}

{$$\sum_{k=1}^{\infty} \left( e^{-{1 \over 2}^{k+k-2}} \right)$$ - $$e^{4 \over 3}$$ - $$e^{16 \over 3}$$ - $$\frac{e}{e-1}$$ - $$\frac{e^{2}}{e-1}$$ + The series diverges.
 * type=""}

Footnote 1
Allow either $$a$$ or $$b$$ to equal something. We allowed $$a$$ to equal something by adding $$b$$ to both sides of the equation.
 * $$a=3+b$$

From there, substitute back into the system of equations, equation (1), to get:
 * $$n+(3+b)=8$$

After, solve for $$n$$ and substitute that into equation (2):
 * (1):$$n+(3+b)-(3+b)=8-(3+b)$$
 * (1):$$n=8-(3+b)=8-3-b=5-b$$
 * (2):$$(5-b)+b=5$$
 * (2):$$5=5$$

In an attempt to solve for $$b$$, the resultant answer tells us an identity (5 always equals 5). Therefore, b has infinitely many solutions. The same thing also happens with $$a$$. Finally, because $$n$$ is equal to itself, the following is also true:
 * $$\begin{cases}

n+a-a=8-a \\ n+b-b=5-b \end{cases}$$
 * $$\begin{cases}

n=8-a \\ n=5-b \end{cases}$$
 * (4): $$8-a=5-b$$.

The following truth reveals itself. As long $$a-b=3$$, equation (3), and $$8-a=5-b$$, equation (4), the following must be true:
 * $$\begin{cases}

a-b=3 \\ 8-a=5-b \end{cases}$$

However, the same situation results when trying to find one solution set $$(a,b)$$: there are infinitely many solutions. We learn one important lesson: what mathematics means by factor is make it so that all terms can be multiplied by another term. There is no specific term you need to care about.$$\blacksquare$$

Physics 1 Problem
{A crate of mass 100 kg rests on the floor. The coefficient of static friction between the floor and the crate is 0.4. If the applied force is 300 N (parallel to the floor), what is true about the system? Select two choices. + The net force is zero on the object. - The crate will slide, thereby having a kinetic friction. - The acceleration of the crate will be $$7\,\tfrac{\text{m}}{\text{s}^{2}}$$. + The crate's magnitude for the force of static friction is $$300\text{N}$$.
 * type="[]"}

Generalizing the Can Optimization
Given the volume $$L$$ of a cylinder. Prove the minimum of the surface area always has $$h=2r$$ for any chosen $$r>0$$. Never thought you would see a proof? Well, here you go.

Small Problems
Only the most simplified answers will be accepted for "gapfill" responses. Use a calculator when needed.

{Use the following situation below to answer items 1 - 4.}

{''A prison determines its price, in dollars, by the number of exports for the license plates per day, $$x$$, according to the demand function $$p(x)=30-0.01x-0.001x^{2}$$. The marginal cost in terms of cost maintaining the materials $$x$$ is $$MC(x)=0.0111111\left(x^{2}-60x+900\right)$$. The cost of maintaining the prison by itself is $200 because of governmental grants and tax money.''}

{Find the marginal revenue of this situation. $$MR(t)=$${ 30 _7 }$$-$${ 0.02|1/50 _7 }$$t-$${ 0.03|3/100 _7 }$$t^{2}$$
 * type="{}" coef="12"}

{At what price should the license plates be sold at to maximize the economic profit of the prison? ${ 15.20|15.2 _7 }
 * type="{}" coef="6"}

{What is the maximized profit of the prison? ${ 162.18 _7 }
 * type="{}" coef="12"}

{What is the quantity demanded from the government for the license plates? ${ 116|117 _5 }
 * type="{}" coef="6"}

{$$B(w)=120w$$ $$p(x)=\frac{1}{2}x^2-50x+1280.5$$}

{A small business wants to decrease the variable cost of baking bread. The variable cost of each corresponding factor of production is shown, whereby $$B(w)$$ represents the variable cost of hiring $$w$$ working bakers at the shop over 10 hours, and $$p(x)$$ represents the variable cost of producing $$x$$ loaves of bread. The fixed cost of this business is $300. The demand function of the firm is perfectly elastic. The marginal revenue is $300 per loaf of bread. Which of the following are true? Select all that apply - The marginal cost here represents a diminishing returns to scale. + The total revenue function is $$R(x)=300x$$. - We are allowed to change the input $$w$$ for $$B(w)$$ and $$x$$ for $$p(x)$$ to an interstitial variable $$u$$ since by adding $$B(w)$$ and $$p(x)$$, we effectively obtain the variable cost as a function of each unit $$u$$ in the factor of productions. - It is impossible to obtain the marginal cost in this situation because the variables of each function above are different, and therefore do not allow us to find the variable cost in this situation. Without the variable cost, the total cost cannot be found, so the marginal cost cannot be found as a result. + The variable cost must involve two inputs as a function $$V(x,w)$$ or the variable cost needs to have at least one of the functions as a constant added to the variable cost. + For 10 workers, the quantity that allows for maximum profit is 350 loaves of bread. The marginal cost in this situation has a diminishing returns to scale.
 * type"[]"}
 * This would be true if the marginal cost can be calculated. Unfortunately, the calculation requires us to ignore one of the factors herein as a constant.
 * Correct. This is because the demand function is perfectly elastic and is therefore a flat line.
 * Because the two functions require different inputs, the variable cost will be different due to a constant for any input. As such, by adding the two functions as if the variable did not matter would greatly obfuscate the actual correct answer as the marginal cost function would be different and therefore lead to incorrect conclusions.
 * The language here is too strong. While it may be impossible to obtain an actual marginal cost, the variable cost can be obtained as a function of two inputs. Because we are trying to find the marginal cost using two inputs, the marginal cost will be different.
 * For any constant of workers $$c_{w}$$, the maximum profit is obtained at 350 loaves of bread because it is simply a constant added to the variable cost. As long as the total revenue does not increase, the quantity required to obtain a maximum profit is 350 loaves of bread.

{A frictionless inclined ramp, angled at $$30^{\circ}$$, of height $$h=10\text{ m}$$, has a $$15\text{ kg}$$ ball at rest at the top. An initial horizontal velocity $$v_{x0}\tfrac{\text{m}}{\text{s}}$$ is applied to the ball. The ball is slowing down along the ramp due to the air resistance of the room, with a counteracting force $$F_{r}=-kv$$ (parallel to the ramp) where $$k$$ is a constant and $$v$$ is the velocity of the moving object. If $$k=2\tfrac{\text{kg}}{\text{s}}$$, what is the work, $$\Delta E$$, done on the system (to the nearest tenth )? Note: Weight is $$F_{g}=mg$$ and $$\Delta E=W=\int_{0}^{b}F(x)dx$$, where $$x$$ is the horizontal position, $$g=9.81\tfrac{\text{m}}{\text{s}^{2}}$$, and $$b$$ is how you arrive at the desired answer. { 3500.7 _7 } $$\text{Joules (J)}$$.
 * type="{}"}