1. **Problem 50(a):** Find an equation connecting total profit $P$ and selling price $x$ given that $P$ partly varies directly as $x$ and partly as $x^2$. 2. Since $P$ varies partly as $x$ and partly as $x^2$, we write: $$P = ax + bx^2$$ where $a$ and $b$ are constants to be determined. 3. Use the given data points: - When $x=20$, $P=60000$: $$60000 = 20a + 400b$$ - When $x=30$, $P=75000$: $$75000 = 30a + 900b$$ 4. Solve the system: From the first equation: $$20a + 400b = 60000$$ From the second: $$30a + 900b = 75000$$ Multiply the first equation by 3: $$60a + 1200b = 180000$$ Multiply the second by 2: $$60a + 1800b = 150000$$ Subtract the first from the second: $$0a + 600b = -30000 \\ b = -50$$ Substitute $b=-50$ into the first equation: $$20a + 400(-50) = 60000 \\ 20a - 20000 = 60000 \\ 20a = 80000 \\ a = 4000$$ 5. The equation connecting $P$ and $x$ is: $$P = 4000x - 50x^2$$ 6. **Problem 50(b):** Find total profit when $x=35$: $$P = 4000(35) - 50(35)^2 = 140000 - 50(1225) = 140000 - 61250 = 78750$$ 7. **Problem 50(c):** Find maximum total profit and corresponding selling price. Since $P = 4000x - 50x^2$ is a quadratic with negative coefficient for $x^2$, it opens downward and has a maximum at vertex: $$x = -\frac{b}{2a} = -\frac{4000}{2(-50)} = \frac{4000}{100} = 40$$ Calculate maximum profit: $$P_{max} = 4000(40) - 50(40)^2 = 160000 - 50(1600) = 160000 - 80000 = 80000$$ --- 8. **Problem 51(a):** The monthly income $y$ is partly constant $k$ and partly varies directly as number of items $x$ sold: $$y = k + mx$$ where $m$ is the rate per item. From the graph, when $x=0$, $y=k$. The graph shows the $y$-intercept at 3000, so: $$k = 3000$$ 9. **Problem 51(b):** Find equation connecting $x$ and $y$. From the graph, pick a point on the line, for example at $x=3$, $y=4500$ (approximate from description). Use: $$4500 = 3000 + 3m \\ 3m = 1500 \\ m = 500$$ So the equation is: $$y = 3000 + 500x$$