1. **Matrix Multiplication Problem:** Given matrices $$P = \begin{bmatrix}-6 & -7 \\ 2 & 0 \\ 5 & q\end{bmatrix}, \quad W = \begin{bmatrix}11 & 9 & -4 \\ -2 & 0 & 1\end{bmatrix}$$ We want to find the product matrix $$PW$$. **Formula:** The element in row $i$, column $j$ of the product matrix is given by $$ (PW)_{ij} = \sum_{k} P_{ik} W_{kj} $$ **Step 1:** Confirm dimensions. Matrix $P$ is $3 \times 2$, matrix $W$ is $2 \times 3$, so product $PW$ is $3 \times 3$. **Step 2:** Calculate each element. - Row 1: - $(PW)_{11} = (-6)(11) + (-7)(-2) = -66 + 14 = -52$ - $(PW)_{12} = (-6)(9) + (-7)(0) = -54 + 0 = -54$ - $(PW)_{13} = (-6)(-4) + (-7)(1) = 24 - 7 = 17$ - Row 2: - $(PW)_{21} = (2)(11) + (0)(-2) = 22 + 0 = 22$ - $(PW)_{22} = (2)(9) + (0)(0) = 18 + 0 = 18$ - $(PW)_{23} = (2)(-4) + (0)(1) = -8 + 0 = -8$ - Row 3: - $(PW)_{31} = (5)(11) + (q)(-2) = 55 - 2q$ - $(PW)_{32} = (5)(9) + (q)(0) = 45 + 0 = 45$ - $(PW)_{33} = (5)(-4) + (q)(1) = -20 + q$ **Result:** $$PW = \begin{bmatrix}-52 & -54 & 17 \\ 22 & 18 & -8 \\ 55 - 2q & 45 & -20 + q\end{bmatrix}$$ --- 2. **Stationary Points and Nature for** $y = 3t^4 - 8t^3 + 6t^2 - 3$ **Step 1:** Find the first derivative to locate stationary points. $$y' = \frac{dy}{dt} = 12t^3 - 24t^2 + 12t$$ **Step 2:** Set $y' = 0$ to find critical points. $$12t^3 - 24t^2 + 12t = 0$$ Divide both sides by 12: $$t^3 - 2t^2 + t = 0$$ Factor: $$t(t^2 - 2t + 1) = 0$$ $$t(t - 1)^2 = 0$$ So stationary points at: $$t = 0, \quad t = 1$$ **Step 3:** Find the second derivative to determine nature. $$y'' = \frac{d^2y}{dt^2} = 36t^2 - 48t + 12$$ Evaluate at $t=0$: $$y''(0) = 12 > 0$$ So $t=0$ is a local minimum. Evaluate at $t=1$: $$y''(1) = 36 - 48 + 12 = 0$$ Second derivative test inconclusive at $t=1$. **Step 4:** Use higher derivative or test values around $t=1$. Third derivative: $$y''' = 72t - 48$$ At $t=1$: $$y'''(1) = 72 - 48 = 24 \neq 0$$ Since $y''(1) = 0$ and $y'''(1) \neq 0$, $t=1$ is a point of inflection. **Summary:** - Stationary points at $t=0$ (local minimum) and $t=1$ (point of inflection). --- 3. **Rate of Change of Sales at $q=1$ for** $$p(q) = \frac{0.3 q^3}{1 + 0.4 q^2}, \quad 0 \leq q \leq 2$$ **Step 1:** Differentiate $p(q)$ using quotient rule: $$p'(q) = \frac{(0.3)(3q^2)(1 + 0.4 q^2) - (0.3 q^3)(0.8 q)}{(1 + 0.4 q^2)^2}$$ Simplify numerator: $$= \frac{0.9 q^2 (1 + 0.4 q^2) - 0.24 q^4}{(1 + 0.4 q^2)^2}$$ $$= \frac{0.9 q^2 + 0.36 q^4 - 0.24 q^4}{(1 + 0.4 q^2)^2} = \frac{0.9 q^2 + 0.12 q^4}{(1 + 0.4 q^2)^2}$$ **Step 2:** Evaluate at $q=1$: $$p'(1) = \frac{0.9 (1)^2 + 0.12 (1)^4}{(1 + 0.4 (1)^2)^2} = \frac{0.9 + 0.12}{(1 + 0.4)^2} = \frac{1.02}{1.4^2} = \frac{1.02}{1.96} \approx 0.5204$$ **Interpretation:** At the beginning of the second year ($q=1$), sales were increasing at approximately 0.5204 million dollars per year.