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: \[ \boxed{3000} \]