Advanced Placement (AP), 26.02.2021 22:10 JPU

Outline the three types of vertical coordination and then describe the key difference between it and vertical integration.

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There are three different types of vertical marketing systems: a corporate system, a contractual system, and an administered system. The three types of vertical marketing systems are contractual, corporate and Administered. A corporate system is desenbed as one in the combination of successive stages of production and distribution under single ownership.

A contractual system is described as an independent production and distribution firms integrate their efforts on a contractual basis to obtain a greater functional economies and marketing impact than they could achieve on their own. There are three variations of contractual systems which include

1.) wholesaler sponsored voluntary chains (wholesaler develops a contractual relationship with a small independent retailer to coordinate and standardize practices)

2.) retailer sponsored (small independent retailers form an organization that operates a wholesale facility cooperatively, for example Ace Hardware company and an individual

3.) Franchising: the most visible variation of contractual systems, it is a contractual agreement between a parent that allows a franchise to operate under an established name, i.e. your family owning a McDonald's

Administered system: achieve coordination at successive stages of production and distribution by the size and influence of one channel member rather than through ownership. For example, Proctor and Gamble as well as Walmart.

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Cause it is

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Answer from: Quest

You need to figure out at which x-value does the graph has a tangent line slope of -1.remember, the derivative is the slope of the tangent line. so find the derivative function and set it to equal -1 to find the x-value that will have a tangent line slope of 0.differentiate with the chain rule. $\begin{aligned} f(x) & = (0.5)^{2x} \\ f'(x) & = 0.5^{2x} \cdot \ln(0.5) \cdot (2x)' \\ f'(x) & = 0.5^{2x} \cdot \ln(0.5) \cdot 2 \\ f'(x) & = 2\ln(0.5) \cdot 0.5^{2x} \end{aligned}$ set the derivative to equal -1 and solve for x. $\begin{aligned} -1 & = 2\ln(0.5) \cdot 0.5^{2x} \\ -\frac{1}{2} & = \ln(0.5) \cdot 0.5^{2x} \\ -\frac{1}{2\ln(0.5)} & = 0.5^{2x} \end{aligned}$ take log base 10 of both sides. also recall that $\log_c(m^n) = n \log_c m$ . $\begin{aligned} \log\left[ -\frac{1}{2\ln(0.5)}\right] & = \log\left[0.5^{2x}\right] \\ \log\left[ -\frac{1}{2\ln(0.5)}\right] & = 2x \log(0.5) \\ \dfrac{\log\left[ -\frac{1}{2\ln(0.5)}\right]}{2\log(0.5)} & = x \\ x & \approx 0.2356 \end{aligned}$ so the x-coord of the point of tangency is $x \approx 0.2356$ .the y-coordinate of this point of tangency can be calculated using the original function: $\begin{aligned} f(x) & = (0.5)^{2x} \\ f(0.2356) & = (0.5)^{2 \cdot 0.2356} \\ & \approx 0.7213 \end{aligned}$ using point-slope form, the equation of the tangent line is $y - 0.7213 = -1(x-0.2356)$ the x-intercept of this tangent line is when the graph is at y=0. $\begin{aligned} 0 - 0.7213 & = -1(x-0.2356)\\ - 0.7213 & = -(x-0.2356) \\ 0.7213 & = x-0.2356\\ x & = 0.9569 \end{aligned}$ this rounds off to $x & = 0.957$ .choice c.

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Outline the three types of vertical coordination and then describe the key difference between it and...

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