1. ## Indefinite Integration

Let f(x) represent the total number of items a subject has memorized x minutes after being presented with a long list of items to learn. Psychologists refer to the graph of y = f(x) as a learning curve and to f'(x) as the learning rate. The time of peak efficiency is the time when the learning rate is maximized. Suppose the learning rate is

$\displaystyle f'(x) = 0.1 (10 + 12x - 0.6x^2)$ for $\displaystyle 0 \leq x \leq 25$

a) When does peak efficiency occur? What is the learning rate of peak efficiency?

b) What is f(x)?

a) Find f(0), f(25)

$\displaystyle 0 = 0.1(10 + 12x - 0.6x^2)$

$\displaystyle 10 + 12x - 0.6x^2 = 0$

$\displaystyle 12x - 0.6x^2 = -10$

$\displaystyle x = 0$

$\displaystyle 1 - 200x = 0$

$\displaystyle 1 = 200x$

$\displaystyle \frac{1}{200} = x$

...but the answer is x = 10 and I don't even know where this number comes from! Plus, the critical points I found are outside the restrictions...

b) $\displaystyle \int 1 \, dx + \int 1.2x \, dx - \int 0.06x^2 \, dx$

$\displaystyle = x + 0.6x^2 - 0.02x^3 + c$

...my question is why is there no "c" (arbitrary constant)??

=====================

A manufacturer estimates marginal revenue to be $\displaystyle 200q^{-\frac{1}{2}}$ dollars per unit when the level of production is q units. The corresponding marginal cost has been found to be 0.4q dollars per unit. If the manufacturer's profit is $2,000 when the level of population is 25 units, what is the profit when the level of production is 36 units?$\displaystyle \int 200q^{-\frac{1}{2}} + 0.4q\, dq\displaystyle = 400q^{\frac{1}{2}} + 0.2q^2 + c\displaystyle 2000 = 400(25)^{\frac{1}{2}} + 0.2(25)^2 + c\displaystyle 2000 = 2125 + c\displaystyle -125 = c\displaystyle R(q) = 400q^{\frac{1}{2}} + 0.2q^2 - 125\displaystyle R(36) = 2534.20$...the right answer is$2265.80, what am I doing wrong?

2. Originally Posted by Macleef
Let f(x) represent the total number of items a subject has memorized x minutes after being presented with a long list of items to learn. Psychologists refer to the graph of y = f(x) as a learning curve and to f'(x) as the learning rate. The time of peak efficiency is the time when the learning rate is maximized. Suppose the learning rate is

$\displaystyle f'(x) = 0.1 (10 + 12x - 0.6x^2)$ for $\displaystyle 0 \leq x \leq 25$

a) When does peak efficiency occur? What is the learning rate of peak efficiency?

b) What is f(x)?

a) Find f(0), f(25)

$\displaystyle 0 = 0.1(10 + 12x - 0.6x^2)$

$\displaystyle 10 + 12x - 0.6x^2 = 0$

$\displaystyle 12x - 0.6x^2 = -10$

$\displaystyle x = 0$

$\displaystyle 1 - 200x = 0$

$\displaystyle 1 = 200x$

$\displaystyle \frac{1}{200} = x$

...but the answer is x = 10 and I don't even know where this number comes from! Plus, the critical points I found are outside the restrictions...
You need to find the maximum in $\displaystyle f'(x)$ so what you need to calculate is the value of x for which $\displaystyle f''(x) = 0$

$\displaystyle f'(x) = 0.1 (10 + 12x - 0.6x^2)$

$\displaystyle f''(x) = 0.1 (12 - 1.2x)$

So solving for $\displaystyle f''(x) = 0$ you get $\displaystyle x=10$ so the only critical point occurs for this value. Now determine if it is a maximum or minimum you need to look at the next derivative, i.e. $\displaystyle f'''(x)$ you will find the sign is negative and therefore x=10 is a global maximum.

The rate of peak efficiency is just

$\displaystyle f'(10) = 0.1 (10 + 12\times 10 - 0.6\times 10^2)$

Originally Posted by Macleef
b) $\displaystyle \int 1 \, dx + \int 1.2x \, dx - \int 0.06x^2 \, dx$

$\displaystyle = x + 0.6x^2 - 0.02x^3 + c$

...my question is why is there no "c" (arbitrary constant)??
One may assume that at time $\displaystyle x=0$ the subject has not memorised anything so

$\displaystyle f(0) = 0$

This can be used to eliminate the constant. ($\displaystyle c=0$).

Originally Posted by Macleef
A manufacturer estimates marginal revenue to be $\displaystyle 200q^{-\frac{1}{2}}$ dollars per unit when the level of production is q units. The corresponding marginal cost has been found to be 0.4q dollars per unit. If the manufacturer's profit is $2,000 when the level of population is 25 units, what is the profit when the level of production is 36 units?$\displaystyle \int 200q^{-\frac{1}{2}} + 0.4q\, dq\displaystyle = 400q^{\frac{1}{2}} + 0.2q^2 + c\displaystyle 2000 = 400(25)^{\frac{1}{2}} + 0.2(25)^2 + c\displaystyle 2000 = 2125 + c\displaystyle -125 = c\displaystyle R(q) = 400q^{\frac{1}{2}} + 0.2q^2 - 125\displaystyle R(36) = 2534.20$...the right answer is$2265.80, what am I doing wrong?
The only thing you have done wrong is to have the wrong sign for the marginal cost. As this is a cost term it should be negative. So start with

$\displaystyle \int 200q^{-\frac{1}{2}} - 0.4q\, dq$

and you should get the correct answer.

Hope this helps.

### a manufacturer has found that marginal cost is (0.4q

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