If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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?
Your solution:
Confidence Rating:
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. **
Your solution:
Confidence Rating:
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?
Your solution:
Confidence Rating:
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 * .08^2 - .08^3) * k = (.0077 - .0016) k = .0061 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):
Self-critique Rating: