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
help needed please.
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
HELP NEEDED PLEASE.
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.
Happy deriving! Don't forget your units in your solutions!