Subjects algebra, calculus, matrix algebra

Matrix Stationary Sales

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Matrix Stationary Sales


1. **Matrix Multiplication Problem:** Given matrices $$P = \begin{bmatrix}-6 & -7 \\ 2 & 0 \\ 5 & 0\end{bmatrix}, \quad W = \begin{bmatrix}11 & 9 & -4 \\ -2 & 0 & 1 \\ 0 & -6 & 1\end{bmatrix}$$ We want to find the product $PW$. Note: Matrix multiplication $AB$ is defined if the number of columns of $A$ equals the number of rows of $B$. Here, $P$ is $3 \times 2$ and $W$ is $3 \times 3$, so $PW$ is not defined because $2 \neq 3$. Therefore, the product $PW$ does not exist. --- 2. **Stationary Points of** $y = 3t^4 - 8t^3 + 6t^2 - 3$: Step 1: Find the derivative $y'$ 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: $$t(t^2 - 2t + 1) = 0$$ Note that $t^2 - 2t + 1 = (t - 1)^2$. So, $$t(t - 1)^2 = 0$$ Step 3: Solve for $t$: $$t = 0 \quad \text{or} \quad t = 1$$ Step 4: Determine the nature of stationary points using the second derivative: $$y'' = \frac{d^2y}{dt^2} = 36t^2 - 48t + 12$$ 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 is inconclusive at $t=1$. Step 5: Use the first derivative test or analyze $y'$ around $t=1$: Since $(t-1)^2$ is squared, the derivative touches zero and does not change sign, indicating a point of inflection at $t=1$. --- 3. **Rate of Change of Sales at Beginning of Second Year:** Given $$p(q) = \frac{0.3 q^3}{1 + 0.4 q^2}, \quad 0 \leq q \leq 2$$ We want $p'(q)$ at $q=1$ (beginning of second year). Step 1: Use quotient rule: $$p'(q) = \frac{(0.9 q^2)(1 + 0.4 q^2) - (0.3 q^3)(0.8 q)}{(1 + 0.4 q^2)^2}$$ Simplify numerator: $$= 0.9 q^2 + 0.36 q^4 - 0.24 q^4 = 0.9 q^2 + 0.12 q^4$$ Step 2: Evaluate at $q=1$: $$p'(1) = \frac{0.9(1)^2 + 0.12(1)^4}{(1 + 0.4(1)^2)^2} = \frac{0.9 + 0.12}{(1 + 0.4)^2} = \frac{1.02}{1.4^2} = \frac{1.02}{1.96} \approx 0.5204$$ Step 3: Interpretation: The sales were increasing at approximately 0.5204 million per unit $q$ at the start of the second year. --- **Final answers:** 1. $PW$ is not defined because matrix dimensions do not match. 2. Stationary points at $t=0$ (local minimum) and $t=1$ (point of inflection). 3. Sales increasing rate at $q=1$ is approximately 0.5204 million per unit $q$.