# Math Help - Most Invested At 9 Percent

1. ## Most Invested At 9 Percent

You are going to invest $30,000, part at 9% and part at 14%. What is the most that can be invested at 9% in order to make at least$3000 in interest per year?

2. You can define a variable.

Let x be the sum invested at 9% interest.
(30000-x) will be the sum invested at 14% interest.

The interest in the first case becomes: $9% \times x = \dfrac{9x}{100}$

The interest in the second case becomes: $(30000-x) \times 14% = \dfrac{14(30000 - x)}{100}$

Total interest = $\dfrac{9x}{100} + \dfrac{14(30000 - x)}{100}$

Now, equate to 3000 to find the value of x, the amount invested to 9% interest.

$\dfrac{9x}{100} + \dfrac{14(30000 - x)}{100} = 3000$

This gives you the greatest value invested at 9% interest to get 3000 interest per year.

If less is invested at 9%, more will be invested at 14% and you get a higher interest.
Ideally, if you want to make the most interest, put everything at 19%

3. After doing the math on paper, I got x = -51,600. I got a negative value. Is this right? I don't think so.

4. You should be getting \$24000

$9x + 14(30000 - x) = 300000$

$9x + 420000 - 14x = 300000$

$-5x + 420000 = 300000$