course Mth 271 026. `query 26*********************************************
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given Solution: `a 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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Given Solution: `a** 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 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). 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 2nd derv is -6 r + .24, which is negative for r > .06. 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.
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:ok "