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Table 2 shows Excalibur Internet's demand table, total revenue, and marginal revenue at each price. Excalibur Internet's marginal cost per internet package is $47.22. What
Table 2 shows Excalibur Internet's demand table, total revenue, and marginal revenue at each price. Excalibur Internet's marginal cost per internet package is $47.22. What is the profit maximizing quantity and price for Excalibur Internet?
Table 2 Demand Schedule and Calculated Revenue
Price | Amount Demanded | Total Revenue | Marginal Revenue |
$160 | 0 | $0 | n/a |
$130 | 125 | $16,250 | $130.00 |
$100 | 250 | $25,000 | $70.00 |
$75 | 475 | $35,625 | $47.22 |
$40 | 625 | $25,000 | -$70.83 |
$0 | 800 | $0 | -$142.86 |
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An electron is incident from a medium with effective mass m onto a medium with effective electron mass m2. The difference in effective mass should cre- ate some fraction reflected and transmit some fraction. Our aim is to find the reflectivity. (a) Using the standard waveforms eikir, re-iki and teik2 with k1,2 be- ing the corresponding electron wave vector values, let us now solve for R = |r|2. The first equation is given by matching wave functions at x = 0. Write down an equation in terms of r and t. Also write down k and k in terms of m, E, Uo, and . [5 pts] (b) In presence of a varying effective mass, the second boundary condition is a bit different from just matching wavefunction derivatives. Just like you derived this for the delta- function potential on your prac- tice problems, let us derive this once again. In presence of a varying effective mass, the Schrodinger equation looks like 1 (x) 2 x m(x) +U(x)(x) = Ev(x) (2) Integrating the equation over a small domain around x = 0, with U(x) = 0 (since the electron is effectively free on both sides, albeit with different masses), write down an equation connecting the deriva- tives. [10 pts] (c) Use the equation connecting the derivatives to get the second equation connecting r and t [5 pts] (d) Solve the two equations to extract the reflectivity R = r. Express this in terms of the electron velocities 1,2 on either side. Recall for a free electron, v= hk/m. [5 + 5 pts] Uo
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