To approximate the total profit when $x = 180$, we compute the sum:

$\sum_{i=1}^6 P'(x_i) \Delta x$

where:

• $\Delta x = 30$,
• $x_i$ are the $x$-values corresponding to the left-hand endpoints of the intervals,
• $P'(x) = -0.0007x^3 + 0.23x^2 + 45.4x$ is the marginal profit.

Step 1: Compute the $x_i$ values

The $x_i$ values correspond to the left endpoints of intervals of width $\Delta x = 30$, starting from 0 up to 150: $x_1 = 0, \quad x_2 = 30, \quad x_3 = 60, \quad x_4 = 90, \quad x_5 = 120, \quad x_6 = 150$

Step 2: Compute $P'(x_i)$ for each $x_i$

Substitute each $x_i$ into $P'(x) = -0.0007x^3 + 0.23x^2 + 45.4x$.

1. $P'(x_1) = P'(0) = -0.0007(0)^3 + 0.23(0)^2 + 45.4(0) = 0$
2. $P'(x_2) = P'(30) = -0.0007(30)^3 + 0.23(30)^2 + 45.4(30)$ $P'(30) = -0.0007(27,000) + 0.23(900) + 45.4(30) = -18.9 + 207 + 1,362 = 1,550.1 \, \text{cents}$
3. $P'(x_3) = P'(60) = -0.0007(60)^3 + 0.23(60)^2 + 45.4(60)$ $P'(60) = -0.0007(216,000) + 0.23(3,600) + 45.4(60) = -151.2 + 828 + 2,724 = 3,400.8 \, \text{cents}$
4. $P'(x_4) = P'(90) = -0.0007(90)^3 + 0.23(90)^2 + 45.4(90)$ $P'(90) = -0.0007(729,000) + 0.23(8,100) + 45.4(90) = -510.3 + 1,863 + 4,086 = 5,438.7 \, \text{cents}$
5. $P'(x_5) = P'(120) = -0.0007(120)^3 + 0.23(120)^2 + 45.4(120)$ $P'(120) = -0.0007(1,728,000) + 0.23(14,400) + 45.4(120) = -1,209.6 + 3,312 + 5,448 = 7,550.4 \, \text{cents}$
6. $P'(x_6) = P'(150) = -0.0007(150)^3 + 0.23(150)^2 + 45.4(150)$ $P'(150) = -0.0007(3,375,000) + 0.23(22,500) + 45.4(150) = -2,362.5 + 5,175 + 6,810 = 9,622.5 \, \text{cents}$

Step 3: Compute the total profit approximation

The total profit is given by: $\text{Profit} = \Delta x \sum_{i=1}^6 P'(x_i)$ Substituting the computed values: $\text{Profit} = 30 \cdot (0 + 1,550.1 + 3,400.8 + 5,438.7 + 7,550.4 + 9,622.5)$ $\text{Profit} = 30 \cdot 27,562.5 = 826,875 \, \text{cents}$

Convert to dollars: $\text{Profit} = \frac{826,875}{100} = 8,268.75 \, \text{dollars}$

Final Answer:

The total profit when 180 members are enrolled is approximately: $\boxed{\$8,268.75}$

Would you like further details or help with related concepts? Here are five related questions:

1. How does the marginal profit function change as $x$ increases?
2. What is the significance of $\Delta x$ in the computation?
3. How would the total profit change if the interval width were reduced (e.g., $\Delta x = 10$)?
4. Can we compute the exact profit using integration instead of summation?
5. How does the cubic term in $P'(x)$ affect the overall marginal profit?

Tip: Always double-check calculations with a calculator or software for precision in applied mathematics problems!