Assignment 26 query

course Mth 271

026. `query 26*********************************************

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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Your solution:

P profit

C = 35x + 500

p = 50 - .1(x)^-1

Revenue = R = price * number sold

R = xp

p = 50 - .1(x)^(-1/2))

R = x(50-.1(x)^(-1/2))

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

Price is revenue – cost:

P = R – C = 50x-.1x^(3/2) – 35x – 500

P = 15x - .1x^(3/2) – 500

P’ = 15 - .1(3/2)x^(1/2)

P’ = 15 – .15x^(1/2) = 0

.15x^(1/2) = 15

x^(1/2) = 100

x = 10000

p = 50 - .1(10,000)^(1/2) = 40

confidence rating:

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I feel fairly confident about this problem.

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Given Solution:

`a Revenue 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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Question:

`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

interest rate = .12

interest rate paid on A is A*r = A*.12

I get confused here, where did the net come from?

The bank pays interest at rate r, so the bank pays the investor A * r.

The bank (note the investor) gets the 12% return.

So the bank gets A * .12, and pays the investor A * r.

Thus the bank nets .12 * A - r * A = (.12 - r) * A.

confidence rating:

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I’m confused about how to finish solving the problem.

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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. **

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Self-critique Rating:

I have not finished solving the problem.

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&#Good responses. See my notes and let me know if you have questions. &#