Solved on Jan 24, 2024

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

STEP 1

Assumptions
1. The total cost function is given by $C(t) = 60 - 40e^{-t}$ .
2. The variable $t$ 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 $t$ .

STEP 2

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

STEP 3

Differentiate the constant term $60$ with respect to $t$ .
$\frac{d}{dt}(60) = 0$

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

STEP 11

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

STEP 12

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

STEP 13

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

STEP 14

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

STEP 15

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

STEP 16

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

STEP 17

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

STEP 18

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

STEP 19

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

STEP 20

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

STEP 21

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

STEP 22

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

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STEP 1

Assumptions1. The total cost function is given by C(t)=60−40e−tC(t) = 60 - 40e^{-t}C(t)=60−40e−t.2. The variable ttt 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 ttt.

STEP 2

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

STEP 3

Differentiate the constant term 606060 with respect to ttt.ddt(60)=0\frac{d}{dt}(60) = 0dtd​(60)=0

STEP 4

Differentiate the exponential term −40e−t-40e^{-t}−40e−t with respect to ttt using the chain rule.ddt(−40e−t)=−40⋅ddt(e−t)\frac{d}{dt}(-40e^{-t}) = -40 \cdot \frac{d}{dt}(e^{-t})dtd​(−40e−t)=−40⋅dtd​(e−t)

STEP 5

Apply the chain rule to differentiate e−te^{-t}e−t.ddt(e−t)=−e−t\frac{d}{dt}(e^{-t}) = -e^{-t}dtd​(e−t)=−e−t

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

Calculate C′(0)C'(0)C′(0) by evaluating the exponential function at t=0t = 0t=0.C′(0)=40e0C'(0) = 40e^{0}C′(0)=40e0

STEP 11

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

STEP 12

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

STEP 13

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

STEP 14

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

STEP 15

STEP 16

As ttt approaches infinity, the term e−te^{-t}e−t approaches zero.lim⁡t→∞e−t=0\lim_{t \rightarrow \infty} e^{-t} = 0t→∞lim​e−t=0

STEP 17

Substitute the limit of e−te^{-t}e−t into the expression for C(t)C(t)C(t).lim⁡t→∞C(t)=60−40⋅0\lim_{t \rightarrow \infty} C(t) = 60 - 40 \cdot 0t→∞lim​C(t)=60−40⋅0

STEP 18

Calculate the limit of C(t)C(t)C(t) as ttt approaches infinity.lim⁡t→∞C(t)=60\lim_{t \rightarrow \infty} C(t) = 60t→∞lim​C(t)=60d) lim⁡t→∞C(t)=60\lim_{t \rightarrow \infty} C(t) = 60limt→∞​C(t)=60

STEP 19

STEP 20

As ttt approaches infinity, the term 40e−t40e^{-t}40e−t approaches zero.lim⁡t→∞40e−t=40⋅lim⁡t→∞e−t\lim_{t \rightarrow \infty} 40e^{-t} = 40 \cdot \lim_{t \rightarrow \infty} e^{-t}t→∞lim​40e−t=40⋅t→∞lim​e−t

STEP 21

Substitute the limit of e−te^{-t}e−t into the expression for C′(t)C'(t)C′(t).lim⁡t→∞C′(t)=40⋅0\lim_{t \rightarrow \infty} C'(t) = 40 \cdot 0t→∞lim​C′(t)=40⋅0

STEP 22

Calculate the limit of C′(t)C'(t)C′(t) as ttt approaches infinity.lim⁡t→∞C′(t)=0\lim_{t \rightarrow \infty} C'(t) = 0t→∞lim​C′(t)=0d) lim⁡t→∞C′(t)=0\lim_{t \rightarrow \infty} C'(t) = 0limt→∞​C′(t)=0

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Assumptions
1. The total cost function is given by $C(t) = 60 - 40e^{-t}$ .
2. The variable $t$ 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 $t$ .

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

Differentiate the constant term $60$ with respect to $t$ .
$\frac{d}{dt}(60) = 0$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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