Let x represent the amount in the fund that pays 6 % interest and y represent the amount in the fund that pays 10 % interest. Write a system of inequalities and solve it by graphing. The minimum amount to be invested in the 10 % fund is the corner of the solution with the smallest y-coordinate.
$3500
Practice makes perfect
Let x represent the amount in the fund that pays 6 % interest and y represent the amount in the fund that pays 10 % interest. Since Mr. Hoffman is investing $10 000 in two funds, the total amount of money will be less than or equal to $10 000.
x+ y ≤ 10 000
We will write another inequality for the interest. To do that we will make an organized table. While writing the inequality, we might want to rewrite the percentages as decimals.
Verbal Expression
Algebraic Expression
Earning from the first fund ($)
0.06 x
Earning from the second fund ($)
0.1 y
Total earning is at least $740.
0.06 x+ 0.1 y ≥ 740
Now we have to inequalities to write a system.
x+y≤ 10 000 & (I) 0.06x+0.1y ≥ 740 & (II)
We will graph the system starting from Inequality I. We will first determine the boundary line of it.
&Inequality I && Boundary Line I
&x+y ≤ 10 000 &&x+y = 10 000
The boundary line is in standard form. Therefore, it would be a better choice to find its intercepts to graph it. We will substitute y= 0 for the x-intercept and x= 0 for the y-intercept.
x+y=10 000
Operation
x-intercept
y-intercept
Substitution
x+ 0=10 000
0+y=10 000
Calculation
x=10 000
y=10 000
Point
(10 000,0)
(0,10 000)
Now we can plot the intercepts and connect them with a line segment. Notice that the amount invested cannot be negative, so it will be bound by the axes. The boundary line will also be solid because of the non-strict inequality.
Next, we will decide which region we should shade. Therefore, we will test the point (0,0).
Since the point satisfies the inequality, region that contains the point will be shaded. Let's do it!
The first inequality has been graphed. Now we will graph the second inequality proceeding in the same way.
&Inequality II && Boundary Line II
&0.06x+0.1y ≥ 740 &&0.06x+0.1y = 740
Next, we will find its intercepts.
0.06x+0.1y = 740
Operation
x-intercept
y-intercept
Substitution
0.06x+0.1( 0) = 740
0.06( 0)+0.1y = 740
Calculation
x=12 333
y=7400
Point
(12 333,0)
(0,7400)
Now that we have found the intercepts, we will test the point (0,0).
The point did not satisfy the inequality, so we will shade the region that does not contain the point.
The overlapping section of the graph above is the solution of the system.
The minimum amount to be invested in the 10 % fund is the corner of the solution with the smallest y-coordinate. In this case, the corner will be the point of intersection of the boundary lines. Let's find it!
