Let's solve each part step by step: --- ### (a) The FrezMore Company #### Given: - Cost function: \( C(x) = 2x^2 + 15x + 1500 \) - Price-demand function: \( p(x) = -0.3x + 460 \) - \( x \) = number of refrigerators (per week), \( 0 \leq x \leq 200 \) #### (i) **Revenue function \( R(x) \):** Revenue = (price per unit) × (number of units sold) \[ R(x) = p(x) \cdot x = (-0.3x + 460)x = -0.3x^2 + 460x \] #### (ii) **Profit function \( P(x) \):** \[ P(x) = R(x) - C(x) \] Plug the expressions: \[ P(x) = [-0.3x^2 + 460x] - [2x^2 + 15x + 1500] \] \[ = -0.3x^2 + 460x - 2x^2 - 15x - 1500 \] \[ = (-0.3 - 2)x^2 + (460 - 15)x - 1500 \] \[ = -2.3x^2 + 445x - 1500 \] #### (iii) **Marginal profit function \( MP(x) \):** Marginal profit is the derivative of the profit function: \[ MP(x) = P'(x) = \frac{d}{dx}[-2.3x^2 + 445x - 1500] \] \[ = -4.6x + 445 \] #### (iv) **Compute \( MP(50) \) and interpret:** \[ MP(50) = -4.6(50) + 445 = -230 + 445 = 215 \] **Interpretation:** When 50 refrigerators are produced, the profit from producing one more refrigerator (the 51st) is approximately \$215. --- ### (b) Investigate the behaviour of \( f(x) \): \[ f(x) = \begin{cases} x & \text{if } x > 0 \\ x+1 & \text{if } x \leq 0 \end{cases} \] - For \( x > 0 \), \( f(x) = x \) - For \( x \leq 0 \), \( f(x) = x + 1 \) As \( x \to 0^+ \), \( f(x) \to 0 \) As \( x \to 0^- \), \( f(x) \to 1 \) So, there is a jump discontinuity at \( x = 0 \). **Graph:** For \( x > 0 \), plot \( y = x \). For \( x \leq 0 \), plot \( y = x + 1 \). At \( x = 0 \), \( f(0) = 1 \) (solid dot), and as you approach from the right it's 0 (open circle at (0,0)). --- ### (c) Sales function: \[ S(t) = 4000 - 4000e^{-0.25t} \] #### (i) **Units sold after 2 months:** \[ S(2) = 4000 - 4000e^{-0.25 \times 2} = 4000 - 4000e^{-0.5} \] Calculate \( e^{-0.5} \approx 0.6065 \): \[ S(2) = 4000 - 4000 \times 0.6065 = 4000 - 2426 = 1574 \] **Final Answer:** About **1574 units** sold after 2 months. #### (ii) **Limit as \( t \to \infty \):** \[ \lim_{t \to \infty} S(t) = 4000 - 4000e^{-0.25t} \] As \( t \to \infty \), \( e^{-0.25t} \to 0 \): \[ S(\infty) = 4000 - 0 = 4000 \] **Interpretation:** The total number of units sold will approach 4000 in the long run. --- ### (d) Evaluate the integral: \[ \int \frac{x^3 + 5x^2 - 4}{\sqrt{x}} dx \] Rewrite numerator terms with powers of \( x \): \[ = \int \left( x^3 / x^{1/2} + 5x^2 / x^{1/2} - 4 / x^{1/2} \right) dx \] \[ = \int \left( x^{5/2} + 5x^{3/2} - 4x^{-1/2} \right) dx \] Integrate term by term: - \( \int x^{5/2} dx = \frac{x^{7/2}}{7/2} = \frac{2}{7} x^{7/2} \) - \( \int 5x^{3/2} dx = 5 \cdot \frac{x^{5/2}}{5/2} = 2x^{5/2} \) - \( \int -4x^{-1/2} dx = -4 \cdot \frac{x^{1/2}}{1/2} = -8x^{1/2} \) **Final Answer:** \[ \boxed{ \int \frac{x^3 + 5x^2 - 4}{\sqrt{x}} dx = \frac{2}{7} x^{7/2} + 2x^{5/2} - 8 x^{1/2} + C } \] --- ## **Summary of Answers** **(a)** - (i) \( R(x) = -0.3x^2 + 460x \) - (ii) \( P(x) = -2.3x^2 + 445x - 1500 \) - (iii) \( MP(x) = -4.6x + 445 \) - (iv) \( MP(50) = 215 \) (Profit increases by \$215 at \( x = 50 \)) **(b)** - Discontinuity at \( x = 0 \), graph as described above. **(c)** - (i) \( S(2) \approx 1574 \) units after 2 months. - (ii) \( \lim_{t \to \infty} S(t) = 4000 \) (long-term sales cap). **(d)** - \( \int \frac{x^3 + 5x^2 - 4}{\sqrt{x}} dx = \frac{2}{7} x^{7/2} + 2x^{5/2} - 8 x^{1/2} + C \)