To determine how many new band uniforms can be purchased for $600 using the cost function \( C(x) = 0.1x^2 + 2.7x + 40 \), we set up the equation with \( C(x) = 600 \):
\[ 0.1x^2 + 2.7x + 40 = 600 \]
Next, we subtract 600 from both sides:
\[ 0.1x^2 + 2.7x + 40 - 600 = 0 \]
This simplifies to:
\[ 0.1x^2 + 2.7x - 560 = 0 \]
To eliminate the decimal, we can multiply the entire equation by 10:
\[ x^2 + 27x - 5600 = 0 \]
Next, we can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 27 \), and \( c = -5600 \).
Calculating the discriminant \( b^2 - 4ac \):
\[ b^2 = 27^2 = 729 \] \[ 4ac = 4 \cdot 1 \cdot (-5600) = -22400 \] Thus,
\[ b^2 - 4ac = 729 + 22400 = 23129 \]
Now we apply the quadratic formula:
\[ x = \frac{-27 \pm \sqrt{23129}}{2 \cdot 1} \] \[ x = \frac{-27 \pm 152}{2} \]
Calculating the two potential values for \( x \):
- \( x = \frac{-27 + 152}{2} = \frac{125}{2} = 62.5 \)
- \( x = \frac{-27 - 152}{2} = \frac{-179}{2} = -89.5 \) (not feasible since we cannot purchase a negative number of uniforms)
Since we need a whole number, we round \( 62.5 \) to the nearest whole number:
\[ x \approx 63 \]
Thus, the number of new uniforms that can be purchased for $600 is approximately 63.