Question

Kyle plans to invest $7500, part of it at 5% simple interest and the rest at 6% simple interest. What is the maximum amount he can invest at 5% and still be guaranteed at least $420 in interest per year? He can invest at most $ _____.

Answer 1

To find the maximum amount Kyle can invest at 5% while ensuring a total interest of at least $420 from both investments, we can set up the following equations.

Let:
- \( x \) be the amount invested at 5%
- \( 7500 - x \) be the amount invested at 6%

The interest earned from the amount invested at 5% is:
\[
0.05x
\]

The interest earned from the amount invested at 6% is:
\[
0.06(7500 - x)
\]

The total interest from both investments must be at least $420:
\[
0.05x + 0.06(7500 - x) \geq 420
\]

Now we can simplify and solve this inequality:

Expanding the second term:
\[
0.05x + 0.06 \times 7500 - 0.06x \geq 420
\]
\[
0.05x + 450 - 0.06x \geq 420
\]

Combining like terms:
\[
-0.01x + 450 \geq 420
\]

Subtracting 450 from both sides:
\[
-0.01x \geq -30
\]

Dividing by -0.01 (and flipping the inequality sign):
\[
x \leq 3000
\]

Thus, the maximum amount that Kyle can invest at 5% and still guarantee at least $420 in interest per year is:
\[
\boxed{3000}
\]

To confirm, if Kyle invests $3000 at 5%, the interest from that investment would be:
\[
0.05 \times 3000 = 150
\]

He would then invest the remainder:
\[
7500 - 3000 = 4500 \quad \text{(at 6%)}
\]

The interest from the 6% investment would be:
\[
0.06 \times 4500 = 270
\]

Thus, the total interest would be:
\[
150 + 270 = 420
\]

This meets the requirement, confirming that investing at most $3000 at 5% is correct. The final answer is:
\[
\boxed{3000}
\]