course Mth 271 ¡ˆ²Tý¸ÜÍú‰ƒÐÖíÈÞ—‡¬£rwÞÜÜp¤assignment #026
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12:19:11 **** Query 3.5.12 find the price per unit p for maximum profit P if C = 35x+500, p=50-.1`sqrt(x) **** What price per unit produces the maximum profit?
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RESPONSE --> P = R - C I know that R = xp so I substitute this into the primary equation. P = xp - C P = x(50 - .1`sqrt(x)) - (35x + 500) P = 50x - .1x^1.5 - 35x - 500 P = -.1x^1.5 + 15x - 500 Now I find the critical point P' = -.15`sqrt(x) + 15 -.15`sqrt(x) + 15 = 0 -.15`sqrt(x) = -15 `sqrt(x) = 100 I can substitute this into the original price function. p = 50 - .1`sqrt(x) p = 50 - .1(100) p = 50 - 10 p = 40 The price per unit that gives the maximum profit is p = 40. confidence assessment: 2
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12:21:41 Revenus is price * number sold: R = xp. Since p = 50 - .1 sqrt(x) we have R = x(50 - .1 `sqrt (x)) = 50x - .1x^(3/2) Price is revenue - cost: P = R - C = 50x - .1 x^(3/2) - 35x - 500. Simplifying: P = 15x - .1x^(3/2) - 500 Derivative of profit P is P ' = 15 -.15 x^(1/2). Derivative is zero when 15 - .15 x^(1/2) = 0; solving we get x = 10,000. 2d derivative is .075 x^-(1/2), which is negative, implying that x = 10000 gives a max. When x = 10,000 we get price p = 50 - .1 sqrt(x) = 50 - .1 * sqrt(10,000) = 40. Price is $40. **
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RESPONSE --> I forgot to test to make sure p=40 was a maximum. I also didn't completely solve for x because I saw that the `sqrt(x) was in the original price function and solving the rest of the way for x would just be adding unnecessary steps. self critique assessment: 2
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12:41:19 ** Query 3.5.22 amount deposited proportional to square of interest rate; bank can reinvest at 12%. What interest rate maximizes the bank's profit? **** What is the desired interest rate?
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RESPONSE --> I'm not quite sure how to do this problem but I gave it a try. Since the amount deposited is proportional to the square of the interest rate I let d = amount deposited and r = interest rate and I wrote d = `sqrt(r) d^2 = r Since P = R - C I thought R would equal the reinvested rate times the amount deposited so R = .12d. I also thought that the cost would be the interest paid. C = r P = .12d - r P = .12d - d^2 P' = .12 - 2d .12 - 2d = 0 -2d = -.12 d = .24 I substituted this into d = `sqrt(r) .24 = `sqrt(r) r = .0576 self critique assessment: 1
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12:45:24 According to my note here amount deposited A is proportional to the square of interest rate r so A = k r^2 for some proportionality constant k. The interest paid at rate r on amount A is A * r. The bank can reinvest at 12% so it gets return A * .12. The bank therefore nets .12 * A - r * A = (.12 - r) * A. Since A = k r^2 the bank nets profit P = (.12 - r) * (k r^2) = k * (.12 r^2 - r^3). We maximize this expression with respect to r: dP/dr = k * (.24 r - 3 r^2). dP/dr = 0 when .24 r - 3 r^2 = 0, when 3 r ( .08 - r) = 0, i.e., when r = 0 or r = .08. The second derivative is -6 r + .24, which is negative for r > .06. This shows that the critical point at r = .08 is a maximum. The max profit is thus P = (.12 * .08 - .08^3) * k = (.096 - .0016) k = .080 k. In order to find the optimal interest rate it is not necessary to find the proportionality constant k. However if the proportionality constant was known we could find the max profit. **
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RESPONSE --> I understand. I messed up and used `sqrt(r) instead of r^2. I also needed a constant k in the proportionality. I should have multiplied the interest paid by the amount deposited also. self critique assessment: 2
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