Solved on Jan 24, 2024

A company's total cost, C(t)=6040etC(t)=60-40 e^{-t}, where tt is time in years. Find: a) C(t)C'(t), b) C(0)C'(0), c) C(6)C'(6), d) limtC(t)\lim_{t\to\infty} C(t) and limtC(t)\lim_{t\to\infty} C'(t).

STEP 1

Assumptions
1. The total cost function is given by C(t)=6040etC(t) = 60 - 40e^{-t}.
2. The variable tt represents the time in years since the start-up date.
3. We need to find the marginal cost, which is the derivative of the total cost function with respect to time tt.

STEP 2

To find the marginal cost, C(t)C'(t), we need to differentiate the total cost function C(t)C(t) with respect to tt.
C(t)=ddt(6040et)C'(t) = \frac{d}{dt}(60 - 40e^{-t})

STEP 3

Differentiate the constant term 6060 with respect to tt.
ddt(60)=0\frac{d}{dt}(60) = 0

STEP 4

Differentiate the exponential term 40et-40e^{-t} with respect to tt using the chain rule.
ddt(40et)=40ddt(et)\frac{d}{dt}(-40e^{-t}) = -40 \cdot \frac{d}{dt}(e^{-t})

STEP 5

Apply the chain rule to differentiate ete^{-t}.
ddt(et)=et\frac{d}{dt}(e^{-t}) = -e^{-t}

STEP 6

Combine the results from STEP_3 and STEP_5 to find C(t)C'(t).
C(t)=0+(40)(et)C'(t) = 0 + (-40)(-e^{-t})

STEP 7

Simplify the expression for C(t)C'(t).
C(t)=40etC'(t) = 40e^{-t}

STEP 8

Now we have the marginal cost function C(t)=40etC'(t) = 40e^{-t}.
a) C(t)=40etC'(t) = 40e^{-t}

STEP 9

To find C(0)C'(0), we substitute t=0t = 0 into the marginal cost function C(t)C'(t).
C(0)=40e0C'(0) = 40e^{-0}

STEP 10

Calculate C(0)C'(0) by evaluating the exponential function at t=0t = 0.
C(0)=40e0C'(0) = 40e^{0}

STEP 11

Since e0=1e^{0} = 1, we can simplify the expression.
C(0)=401C'(0) = 40 \cdot 1

STEP 12

Calculate the value of C(0)C'(0).
C(0)=40C'(0) = 40
b) C(0)=40C'(0) = 40

STEP 13

To find C(6)C'(6), we substitute t=6t = 6 into the marginal cost function C(t)C'(t).
C(6)=40e6C'(6) = 40e^{-6}

STEP 14

Calculate C(6)C'(6) by evaluating the exponential function at t=6t = 6.
C(6)=40e6C'(6) = 40e^{-6}
c) C(6)=40e6C'(6) = 40e^{-6}

STEP 15

To find the limit of C(t)C(t) as tt approaches infinity, we evaluate limtC(t)\lim_{t \rightarrow \infty} C(t).
limtC(t)=limt(6040et)\lim_{t \rightarrow \infty} C(t) = \lim_{t \rightarrow \infty} (60 - 40e^{-t})

STEP 16

As tt approaches infinity, the term ete^{-t} approaches zero.
limtet=0\lim_{t \rightarrow \infty} e^{-t} = 0

STEP 17

Substitute the limit of ete^{-t} into the expression for C(t)C(t).
limtC(t)=60400\lim_{t \rightarrow \infty} C(t) = 60 - 40 \cdot 0

STEP 18

Calculate the limit of C(t)C(t) as tt approaches infinity.
limtC(t)=60\lim_{t \rightarrow \infty} C(t) = 60
d) limtC(t)=60\lim_{t \rightarrow \infty} C(t) = 60

STEP 19

To find the limit of C(t)C'(t) as tt approaches infinity, we evaluate limtC(t)\lim_{t \rightarrow \infty} C'(t).
limtC(t)=limt40et\lim_{t \rightarrow \infty} C'(t) = \lim_{t \rightarrow \infty} 40e^{-t}

STEP 20

As tt approaches infinity, the term 40et40e^{-t} approaches zero.
limt40et=40limtet\lim_{t \rightarrow \infty} 40e^{-t} = 40 \cdot \lim_{t \rightarrow \infty} e^{-t}

STEP 21

Substitute the limit of ete^{-t} into the expression for C(t)C'(t).
limtC(t)=400\lim_{t \rightarrow \infty} C'(t) = 40 \cdot 0

STEP 22

Calculate the limit of C(t)C'(t) as tt approaches infinity.
limtC(t)=0\lim_{t \rightarrow \infty} C'(t) = 0
d) limtC(t)=0\lim_{t \rightarrow \infty} C'(t) = 0

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