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.

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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.
To maximize annual interest, invest \$10 million in Treasury bonds and \$35 million in mutual funds, resulting in a maximum annual interest of \$2.05 million.

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