Problem Overview:

• Marginal annual savings: $S'(x) = 225 - x^2$
• Marginal annual costs: $C'(x) = x^2 + 7x$
• Net savings: $N(x) = \int (S'(x) - C'(x)) \, dx$

Steps to Solve:

1. Maximize net savings $N'(x) = S'(x) - C'(x)$ by finding critical points and testing for maxima.
2. Compute the net savings during the first year of use ($N(1)$).
3. Compute total net savings over the optimal usage period.

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(a) To maximize net savings:

We maximize the difference of the marginal savings and marginal costs: $N'(x) = S'(x) - C'(x) = (225 - x^2) - (x^2 + 7x) = 225 - 2x^2 - 7x$

Set $N'(x) = 0$ to find critical points: $225 - 2x^2 - 7x = 0$

Simplify: $2x^2 + 7x - 225 = 0$

Solve the quadratic equation: $x = \frac{-7 \pm \sqrt{7^2 - 4(2)(-225)}}{2(2)}$

Substitute values: $x = \frac{-7 \pm \sqrt{49 + 1800}}{4} = \frac{-7 \pm \sqrt{1849}}{4} = \frac{-7 \pm 43}{4}$

Compute roots: $x = \frac{-7 + 43}{4} = 9, \quad x = \frac{-7 - 43}{4} = -12.5 \quad (\text{discard negative root})$

Thus, the company should use the machine for 9 years to maximize net savings.

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(b) Net savings during the first year of use:

To compute the net savings during the first year, evaluate: $N(1) = \int_0^1 \left( S'(x) - C'(x) \right) \, dx$

Substitute $S'(x) - C'(x)$: $N(1) = \int_0^1 \left( 225 - 2x^2 - 7x \right) \, dx$

Split the integral: $N(1) = \int_0^1 225 \, dx - \int_0^1 2x^2 \, dx - \int_0^1 7x \, dx$

Solve each term:

1. $\int_0^1 225 \, dx = 225[x]_0^1 = 225$
2. $\int_0^1 2x^2 \, dx = 2 \cdot \frac{x^3}{3} \Big|_0^1 = \frac{2}{3}$
3. $\int_0^1 7x \, dx = 7 \cdot \frac{x^2}{2} \Big|_0^1 = \frac{7}{2}$

Combine results: $N(1) = 225 - \frac{2}{3} - \frac{7}{2}$

Convert to a common denominator: $N(1) = 225 - \frac{4}{6} - \frac{21}{6} = 225 - \frac{25}{6}$

Simplify: $N(1) = \frac{1350}{6} - \frac{25}{6} = \frac{1325}{6} \approx 220.83$

The net savings during the first year is approximately $220.83.

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(c) Total net savings over 9 years:

The total net savings is: $N(9) = \int_0^9 \left( 225 - 2x^2 - 7x \right) \, dx$

Solve: $N(9) = \int_0^9 225 \, dx - \int_0^9 2x^2 \, dx - \int_0^9 7x \, dx$

1. $\int_0^9 225 \, dx = 225[x]_0^9 = 225 \cdot 9 = 2025$
2. $\int_0^9 2x^2 \, dx = 2 \cdot \frac{x^3}{3} \Big|_0^9 = \frac{2}{3} \cdot 9^3 = \frac{2}{3} \cdot 729 = 486$
3. $\int_0^9 7x \, dx = 7 \cdot \frac{x^2}{2} \Big|_0^9 = 7 \cdot \frac{9^2}{2} = 7 \cdot 40.5 = 283.5$

Combine results: $N(9) = 2025 - 486 - 283.5 = 1255.5$

The total net savings over 9 years is $1255.50.

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Summary of Results:

(a) The machine should be used for 9 years to maximize net savings.
(b) Net savings during the first year is approximately $220.83.
(c) Total net savings over the optimal 9-year period is $1255.50.

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Questions for Deeper Understanding:

1. How do marginal savings and costs compare over time, and what does this imply for machine usage?
2. What happens to net savings if the machine is used beyond 9 years?
3. How does increasing $S'(x)$ or decreasing $C'(x)$ impact the results?
4. Can this method be adapted to model different cost or savings functions?
5. How sensitive is the solution to changes in cost or savings parameters?

Tip: Always double-check the setup of equations when working with marginal functions to ensure proper integration limits and accuracy.