#$&* course Mth 271 Question: `q **** 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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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 $40. ** `a** Query 3.5.22 amount deposited proportional to square of interest rate; bank can reinvest at 12%. What interest rate maximizes the bank's profit? **** What is the desired interest rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** STUDENT QUESTION I understand why and how you are taking the derivative and finding the critical numbers , but I'm not sure about where you obtained the formulas and tied everything together???? INSTRUCTOR RESPONSE You might also want to review the modeling project on power functions and proportionality. To say that y is proportional to x is to say that there exists a constant k such that y = k x. Therefore to say that the amount deposited is proportional to the square of the interest rate is to say that A = k * r^2. The rest of the solution follows from that. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Again, it all makes perfect sense after I read it, but remembering how to set it all up is a hard part. ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!