query26

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course Mth 271

11/28 9p

uestion: `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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Your solution:

p = 50-.1(sq. rt. of x) P = R-C

P = x(50-.1(sq. rt. of x) )- (35x+500)

= 50x - .1x^(3/2) - 35x -500

= 15x-.1x^(3/2) - 500 P’ = 15-(3/2)(.1)x^(½)

= 15- .15/(sq. rt. of x) = 0

x^(½) = 100

x= 10,000

p = 50-.1(sq. rt. of 10000)

= 50-.1(100)

= 50-10

= 40

$40 per unit for maximum profit

confidence rating #$&*:

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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?

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Your solution:

A= kr^2

P = (.12-r)(kr^2) = k(.12r^2 - r^3) P’ = k(.24r-3r^2) = 04

.24r-3r^2 = 0

3r(.08-r) = 0 .08 = r

k(.12(.08)^2) - (.08)^3

k(.000768) - (.000512) = .000256

I really can’t explain what this number means because it doesn’t make any sense as an interest rate or an addition/subtraction to .08.

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.

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Self-critique (if necessary):

k * (.12 r^2 - r^3). this does not match

The max profit is thus P = (.12 * .08 - .08^3) * k

It seems like it would be (.12*.08^2-.08^3)*k which instead of being a nice neat number is

k( .000768-.000512).

I don’t understand where the square of the .08 went when you substituted.

This was a difficult problem for me because there was no example in the book to follow.

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Again your attention to detail has detected another error, which I've just corrected.

I do apologize for all the errors, but be assured that you continue to gain points for detecting them.

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That error does not affect the solution to the problem, which is that the optimal interest rate is .08, but it certainly would affect the calculation of the bank's profit.

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