To find out how many new band uniforms can be purchased for $600, we need to set the cost model equation \( C(x) = 0.1x^2 + 2.7x + 40 \) equal to \( 600 \) and solve for \( x \):
\[ 0.1x^2 + 2.7x + 40 = 600 \]
Subtract \( 600 \) from both sides:
\[ 0.1x^2 + 2.7x + 40 - 600 = 0 \]
This simplifies to:
\[ 0.1x^2 + 2.7x - 560 = 0 \]
To simplify calculations, we can multiply the entire equation by \( 10 \) to eliminate the decimal:
\[ x^2 + 27x - 5600 = 0 \]
Now, we will solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = 27 \), and \( c = -5600 \).
Calculate \( b^2 - 4ac \):
\[ b^2 - 4ac = 27^2 - 4 \cdot 1 \cdot (-5600) \] \[ = 729 + 22400 \] \[ = 23129 \]
Now compute the square root:
\[ \sqrt{23129} \approx 152.07 \]
Now substitute back into the quadratic formula:
\[ x = \frac{-27 \pm 152.07}{2} \]
Calculate the two possible solutions for \( x \):
- Using the positive root:
\[ x = \frac{-27 + 152.07}{2} \approx \frac{125.07}{2} \approx 62.535 \]
- Use the negative root (but it will yield a negative quantity which isn't valid):
\[ x = \frac{-27 - 152.07}{2} \approx \frac{-179.07}{2} \approx -89.535 \]
Since \( x \) must be a non-negative whole number, we round \( 62.535 \) to the nearest whole number:
The number of new uniforms that can be purchased for $600 is approximately \( \boxed{63} \).