1. **State the problem:** We are given a quadratic profit function $P = ax^2 + bx + c$ where $x$ is the number of items produced and $P$ is the weekly profit. We have three data points: $(50, 1350)$, $(200, 5100)$, and $(400, 3100)$. 2. **Write down three equations using the data points:** Substitute each $(x, P)$ pair into $P = ax^2 + bx + c$: $$ \begin{cases} 1350 = a(50)^2 + b(50) + c \\ 5100 = a(200)^2 + b(200) + c \\ 3100 = a(400)^2 + b(400) + c \end{cases} $$ Which simplifies to: $$ \begin{cases} 1350 = 2500a + 50b + c \\ 5100 = 40000a + 200b + c \\ 3100 = 160000a + 400b + c \end{cases} $$ 3. **Solve for $a$, $b$, and $c$:** Subtract the first equation from the second and third to eliminate $c$: $$ \begin{aligned} (5100 - 1350) &= (40000a - 2500a) + (200b - 50b) + (c - c) \\ 3750 &= 37500a + 150b \\ \Rightarrow 3750 = 37500a + 150b \quad (1) \end{aligned} $$ $$ \begin{aligned} (3100 - 1350) &= (160000a - 2500a) + (400b - 50b) + (c - c) \\ 1750 &= 157500a + 350b \\ \Rightarrow 1750 = 157500a + 350b \quad (2) \end{aligned} $$ Divide equation (1) by 150: $$\frac{3750}{150} = \frac{37500a}{150} + \frac{150b}{150} \Rightarrow 25 = 250a + b \quad (3)$$ Divide equation (2) by 350: $$\frac{1750}{350} = \frac{157500a}{350} + \frac{350b}{350} \Rightarrow 5 = 450a + b \quad (4)$$ Subtract (4) from (3): $$ 25 - 5 = (250a - 450a) + (b - b) \\ 20 = -200a \\ \Rightarrow a = -\frac{20}{200} = -0.1 $$ Substitute $a = -0.1$ into (3): $$ 25 = 250(-0.1) + b \\ 25 = -25 + b \\ \Rightarrow b = 50 $$ Substitute $a$ and $b$ into the first original equation: $$ 1350 = 2500(-0.1) + 50(50) + c \\ 1350 = -250 + 2500 + c \\ 1350 = 2250 + c \\ \Rightarrow c = 1350 - 2250 = -900 $$ 4. **Interpret $c$:** $c$ is the profit when no items are produced ($x=0$). Here, $c = -900$ means a loss of 900 when nothing is produced, likely fixed costs. 5. **Find the number of items to maximize profit:** The vertex of $P = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. $$ x = -\frac{50}{2(-0.1)} = -\frac{50}{-0.2} = 250 $$ 6. **Calculate the maximum profit:** Substitute $x=250$ into $P$: $$ P = -0.1(250)^2 + 50(250) - 900 = -0.1(62500) + 12500 - 900 = -6250 + 12500 - 900 = 4350 $$ **Final answers:** - $a = -0.1$, $b = 50$, $c = -900$ - $c$ represents a fixed loss of 900 when no items are produced. - Maximum profit is 4350 when producing 250 items.