Question Statement

A company models its operational cost as:

$C(t)=500e^{0.04t}-100t$

where $t$ is the time in years.

a. Find the rate of change of cost at any time $t$.

b. Determine when the rate of increase in cost is minimal.

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Background and Explanation

This problem requires calculating derivatives to analyze how a cost function changes over time. You will need to apply the chain rule for exponential functions and use the second derivative to determine the behavior (increasing/decreasing) of the rate of change itself.

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Solution

Part (a)

To find the rate of change of cost at any time $t$, we calculate the first derivative of the cost function $C(t)$ with respect to time.

Given: $C(t)=500e^{0.04t}-100t$

Differentiating with respect to $t$:

• For the term $500e^{0.04t}$, apply the chain rule: $\frac{d}{dt}[e^u]=e^u\cdot \frac{du}{dt}$ where $u=0.04t$, so $\frac{du}{dt}=0.04$
• For the term $-100t$, the derivative is $-100$

$\begin{array}{rl}\frac{dC}{dt}& =500\cdot e^{0.04t}\cdot (0.04)-100\\ & =20e^{0.04t}-100\end{array}$

Therefore, the rate of change of cost at any time $t$ is: $\frac{dC}{dt}=20e^{0.04t}-100$

Part (b)

To determine when the rate of increase (i.e., $\frac{dC}{dt}$) is minimal, we must find the minimum value of the rate function $\frac{dC}{dt}=20e^{0.04t}-100$.

Since this is an optimization problem for the rate function, we examine its derivative (which is the second derivative of the original cost function $C(t)$):

Differentiate $\frac{dC}{dt}$ with respect to $t$: $\begin{array}{rl}\frac{d^{2}C}{d{t}^2}& =\frac{d}{dt}\left[20e^{0.04t}-100\right]\\ & =20\cdot e^{0.04t}\cdot (0.04)-0\\ & =0.8e^{0.04t}\end{array}$

Analyze the sign of the second derivative: Since $e^{0.04t}>0$ for all real values of $t$, we have: $\frac{d^{2}C}{d{t}^2}=0.8e^{0.04t}>0\text{for all }t$

This result indicates that $\frac{dC}{dt}$ is a strictly increasing function for all $t$ (its slope is always positive). Consequently, the rate of change $\frac{dC}{dt}$ achieves its minimum value at the smallest possible value of $t$.

Assuming time begins at $t=0$ (since $t$ represents time in years), the rate of increase in cost is minimal when: $t=0$

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Key Formulas or Methods Used

• Chain Rule: $\frac{d}{dt}\left[e^{u(t)}\right]=e^{u(t)}\cdot u^{\prime }(t)$
• Exponential Differentiation: $\frac{d}{dt}\left[e^{kt}\right]=k{e}^{kt}$
• Power Rule: $\frac{d}{dt}[t^n]=n{t}^{n-1}$
• Second Derivative Test: Used to determine concavity and monotonicity of the first derivative
• Optimization Principle: If $f^{\prime }(x)>0$ for all $x$ in a domain, then $f(x)$ is strictly increasing and its minimum on $[a,\infty )$ occurs at $x=a$

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Summary of Steps

1. Identify the cost function: $C(t)=500e^{0.04t}-100t$
2. Find the first derivative (rate of change): $\frac{dC}{dt}=20e^{0.04t}-100$ using the chain rule
3. Find the second derivative: $\frac{d^{2}C}{d{t}^2}=0.8e^{0.04t}$ to analyze the behavior of the rate
4. Analyze the sign: Note that $\frac{d^{2}C}{d{t}^2}>0$ for all $t$, proving $\frac{dC}{dt}$ is strictly increasing
5. Determine the minimum: Since the rate function is always increasing, its minimum value occurs at the initial time $t=0$