Assignment 26

course Mth 271

ŠՌ~Pʒf~assignment #026

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026. `query 26

Applied Calculus I

04-25-2007

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23:35:06

**** 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 -->

You begin by solving for P=R-C

P=xp-C

P=xp - (35x + 500)

P=x(50-0.1sqrtx) - 35x + 500

P= 50x - 0.1 sqrtx - 35x + 500

P=15x - .1x^3/2 - 500

This gives you the function of one variable.

To maximize the profit you have to find the critical number.

15-.15x^1/2=0

15 - .075x = 0

-.075x = -15

x = 200

Filling in x into the demand function you get:

p=50-0.1sqrtx

p=50-0.1sqrt200

p=50-1.4142

p=48.5858

confidence assessment: 2

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23:37:07

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 had the principle corrrect but the numbers wrong. I don't understand how you got the 10,000.

self critique assessment: 2

Starting with

15 - .15 x^(1/2) = 0 we subtract 15 from both sides and get

-.15 x^(1/2) = -15. We then divide both sides by -.15 to get

x^(1/2) = -15 / (-.15) = 100. We square both sides to get

(x^(1/2))^2 = 100^2 so that

x = 100^2, or 10,000.

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23:39:01

** 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 -->

P= R-C

.12x^2 = .24x

P=xp-C = .12x^2 - C

P=.24x-C=0

This is as far as I could go. I don't really understand where to go from here. This one is challenging.

self critique assessment: 2

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23:40:35

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 have copied your answer and will review it further to understand how to solve this problem. This was a hard one.

self critique assessment: 2

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"

You had the right general approach on the first problem; see my notes for the solution of the critical-value equation.

A detailed self-critique is recommended on the second problem.