A pension fund manager decides to invest a total of at most \( \$ 45 \) million in U.S. Treasury bonds paying \( 3 \% \) annual interest and in mutual funds paying \( 5 \% \) annual interest. She plans to invest at least \( \$ 5 \) million in bonds and at least \( \$ 20 \) million in mutual funds. Bonds have an initial fee of \( \$ 100 \) per million dollars, while the fee for mutual funds is \( \$ 200 \) per million. The fund manager is allowed to spend no more than \( \$ 8000 \) on fees. How much should be invested in each to maximize annual interest? What is the maximum annual interest? The amount that should be invested in Treasury bonds is \( \$ \square \) million and the amount that should be invested in mutual funds is \( \$ \square \) million.

Let \[ x = \text{investment in Treasury bonds (in million dollars)} \] \[ y = \text{investment in mutual funds (in million dollars)} \] We are given the following constraints: 1. The total investment is at most \$45 million: \[ x + y \le 45. \] 2. At least \$5 million must be invested in bonds: \[ x \ge 5. \] 3. At least \$20 million must be invested in mutual funds: \[ y \ge 20. \] 4. The fees are \$100 per million for bonds and \$200 per million for mutual funds, and the total fee cannot exceed \$8000. Writing the fee constraint: \[ 100x + 200y \le 8000. \] Dividing this inequality by 100 gives: \[ x + 2y \le 80. \] The annual interests are 3% for bonds and 5% for mutual funds. The total annual interest (in million dollars) is \[ \text{Interest} = 0.03x + 0.05y. \] Our goal is to maximize \[ Z = 0.03x + 0.05y, \] subject to: \[ \begin{cases} x + y \le 45, \\ x + 2y \le 80, \\ x \ge 5, \\ y \ge 20. \end{cases} \] We now find the vertices of the feasible region. 1. **Intersection of \( x+y=45 \) and \( x+2y=80 \):** Solve: \[ \begin{cases} x+y = 45, \\ x+2y = 80. \end{cases} \] Subtract the first from the second: \[ (x+2y) - (x+y) = 80-45 \quad \Longrightarrow \quad y = 35. \] Then, \[ x = 45 - y = 45 - 35 = 10. \] So one vertex is \((10, 35)\). 2. **Intersection of \( x+y = 45 \) with \( y = 20 \):** Set \( y = 20 \) in \( x+y = 45 \): \[ x + 20 = 45 \quad \Longrightarrow \quad x = 25. \] So another vertex is \((25,20)\). 3. **Intersection of \( x+2y = 80 \) with \( x = 5 \):** Set \( x = 5 \) in \( x+2y=80 \): \[ 5 + 2y = 80 \quad \Longrightarrow \quad 2y = 75 \quad \Longrightarrow \quad y = 37.5. \] So a vertex is \((5, 37.5)\). We also consider the lower bounds \( x = 5 \) and \( y = 20 \). The point \((5, 20)\) is feasible, but we will check the interest at the vertices found above since the optimum for a linear programming problem occurs at one of the vertices. We now calculate the annual interest \( Z \) at these vertices: - At \((10,35)\): \[ Z = 0.03(10)+0.05(35)=0.3+1.75=2.05. \] - At \((25,20)\): \[ Z = 0.03(25)+0.05(20)=0.75+1.00=1.75. \] - At \((5,37.5)\): \[ Z = 0.03(5)+0.05(37.5)=0.15+1.875=2.025. \] - At \((5,20)\): \[ Z = 0.03(5)+0.05(20)=0.15+1.00=1.15. \] The maximum annual interest is achieved at \((10,35)\) with \[ Z=2.05 \text{ million dollars}. \] Thus, the optimum investment is: - Treasury bonds: \( \$10 \) million, - Mutual funds: \( \$35 \) million, with a maximum annual interest of \( \$2.05 \) million.