Let's solve the given problems step by step for the sales rate $s(t) = 1035t e^{-t^2 / 4}$, where $t$ is time in years.

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a. Total Sales Over 10 Years

The total sales over 10 years can be calculated by integrating the sales rate $s(t)$ from $t = 0$ to $t = 10$:

$\text{Total Sales} = \int_{0}^{10} 1035t e^{-t^2 / 4} \, dt$

To compute this integral:

1. Recognize this as a Gaussian-like integral that requires substitution.
2. Use substitution $u = t^2 / 4$, then $du = t/2 \, dt$.
3. Update limits accordingly and evaluate.

I will calculate this numerically for precision.

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b. When Does the Sales Rate Drop to 10.35?

To find when the sales rate drops to $s(t) = 10.35$, solve the equation:

$1035t e^{-t^2 / 4} = 10.35$

Simplify:

$t e^{-t^2 / 4} = 0.01$

This can be solved numerically using iterative methods or graphing, as it involves a transcendental equation.

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c. Maximum Sales Rate

The maximum sales rate occurs where the derivative of $s(t)$ is zero:

$s'(t) = \frac{d}{dt}\left( 1035t e^{-t^2 / 4} \right)$

Use the product rule: $s'(t) = 1035 \left( e^{-t^2 / 4} - \frac{t^2}{4} e^{-t^2 / 4} \right)$

Factorize: $s'(t) = 1035 e^{-t^2 / 4} \left( 1 - \frac{t^2}{4} \right)$

Set $s'(t) = 0$:

$1 - \frac{t^2}{4} = 0$

Solve for $t$: $t = 2$

The maximum sales rate occurs at $t = 2$. Substituting $t = 2$ into $s(t)$: $s(2) = 1035 \cdot 2 \cdot e^{-2^2 / 4} = 1035 \cdot 2 \cdot e^{-1}$

I will compute this numerically for the exact value.

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Let me calculate these results np from scipy.integrate import quad from scipy.optimize import fsolve

Define the sales rate function

def s(t): return 1035 * t * np.exp(-t**2 / 4)

a. Total sales over 10 years

total_sales, _ = quad(s, 0, 10)

b. Solve for when sales rate drops to 10.35

def sales_rate_equation(t): return 1035 * t * np.exp(-t**2 / 4) - 10.35

time_to_drop = fsolve(sales_rate_equation, 1)[0] # Initial guess of 1 year

c. Maximum sales rate

Derivative of sales rate

def s_prime(t): return 1035 * np.exp(-t2 / 4) * (1 - t2 / 4)

Solve for critical point where s'(t) = 0

time_of_max = fsolve(lambda t: s_prime(t), 2)[0] # Initial guess of 2 years max_sales_rate = s(time_of_max)

total_sales, time_to_drop, time_of### Results:

a. Total Sales Over 10 Years:
The total sales over 10 years is approximately 2070 units.

b. When Does the Sales Rate Drop to 10.35?
The sales rate drops to $10.35$ approximately at $t = 0.01$ years (around 3.65 days).

c. Maximum Sales Rate:
The maximum sales rate occurs at $t = 2$ years, with a sales rate of approximately 761.51 units per year.

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Follow-Up Questions:

1. Would you like to see the detailed steps for solving the integral in part (a)?
2. Should I provide a graph of $s(t)$ over time to visualize these results?
3. Would you like more explanation about how we solved the transcendental equation in part (b)?
4. Should we explore the sensitivity of sales beyond 10 years?
5. Do you need a derivation for finding critical points of $s(t)$ in part (c)?

Tip: When analyzing functions involving $e^{-x^2}$, numerical methods or approximations often provide practical solutions when closed-form answers are not feasible.