1. ## Differentiation/rate of chaange

A certain industrial machine depreciates so that its value t years after purchase is given by
Q(t) = 75,000 e -0.05t

i) What was its value at the time of purchase?

ii) What will its value be 10 years after purchase?
iii) At what rate is the value of the machine depreciating after t years? Hint: differentiate Q(t).

iv) Does this rate depend on t or is it a constant?

v) At what rate is the value of the machine depreciating after 10 years

2. Originally Posted by shizelle

i) What was its value at the time of purchase?
make $\displaystyle t = 0$

Originally Posted by shizelle

ii) What will its value be 10 years after purchase?
make $\displaystyle t = 10$

3. ## answer to question part 1

Q(0)=75000 e^-0.05*0
=75000

4. Originally Posted by shizelle
Q(0)=75000 e^-0.05*0
=75000

Now do the second one

$\displaystyle 75000e^{-0.05\times 10}= \dots$

5. ## part 2 solution

Q(10)=75000 e^-0.5
=45489.8

6. ## Differentiation & integration

A manufacturer finds that the total cost of producing a product is given by the cost function c(q) = 0.05q2 + 5q + 500. At what level of output will average cost per unit be a minimum?

Evaluate the integral below between the limits of 1 and 10

( vx + 3 x -1 2ex) dx

7. Originally Posted by shizelle

Evaluate the integral below between the limits of 1 and 10

( vx + 3 x -1 2ex) dx
Is this the integral?

$\displaystyle \int_0^{10} vx+3x-1-2e^x~dx$

You should start new questions in a separate thread.

8. I am not sure, that's how they gave us the question but I believe it is suppposed to be that way. I cant get through to posting a new thread.

9. I'm gonna guess that -.05t is the exponent. Based on that assumption, the value at the time of purchase, in other words--at t=0, is 75000e^-.05x0, which is 75000, because .05 times 0 = 0, and anything to the xero power is at t=10, you just plug 7500o(e^-.05*10) into your calculator.

the derivative of any function gives the function's instantaneous rate of change at a point. SO to find the rate of depreciation at time t, meaning to find an equation that will give instantaneous rate of change at any value of t, find d/dt(f(x)), or the first derivative of the equation. That is -3750e^(-.05t).

You can logically determine that the rate of change depends on t, because t is in the exponent. If it were constant the equation would be constant , like y=2, or linear, like y=mx+b, right?

Okie dokie. Plug 10 in for t in the first derivative equation, and you'll get you're rate of change at t=10.

10. Originally Posted by pickslides
Is this the integral?

$\displaystyle \int_0^{10} vx+3x-1-2e^x~dx$

You should start new questions in a separate thread.
Yes that is how it is supposed to be, i guess.