Suppose the average cost to community hospitals per patient per day in year \( x \) is approximated by \( \frac{72.469 x^{2}+1353.5 x+2136 s}{x+4} \), where \( x=11 \) comtesponds te the year 2 Will the average cost per petient per day reach \( \$ 4000 \) by 2024 ? What is the value of \( x \) in this case? \( x=\square \)

To determine whether the average cost per patient per day will reach \$4,000 by 2024, we'll solve for the value of \( x \) in the given cost equation: \[ \frac{72.469x^2 + 1353.5x + 2136}{x + 4} = 4000 \] ### Step-by-Step Solution: 1. **Set Up the Equation:** \[ \frac{72.469x^2 + 1353.5x + 2136}{x + 4} = 4000 \] 2. **Multiply Both Sides by \( x + 4 \) to Eliminate the Denominator:** \[ 72.469x^2 + 1353.5x + 2136 = 4000(x + 4) \] 3. **Expand the Right Side:** \[ 72.469x^2 + 1353.5x + 2136 = 4000x + 16000 \] 4. **Bring All Terms to One Side to Form a Quadratic Equation:** \[ 72.469x^2 + 1353.5x + 2136 - 4000x - 16000 = 0 \] \[ 72.469x^2 - 2646.5x - 13864 = 0 \] 5. **Apply the Quadratic Formula:** \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where: - \( a = 72.469 \) - \( b = -2646.5 \) - \( c = -13864 \) 6. **Calculate the Discriminant (\( D \)):** \[ D = (-2646.5)^2 - 4 \times 72.469 \times (-13864) = 11,022,803.114 \] 7. **Find the Square Root of the Discriminant:** \[ \sqrt{D} \approx 3320.06 \] 8. **Solve for \( x \):** \[ x = \frac{2646.5 + 3320.06}{2 \times 72.469} \approx \frac{5966.56}{144.938} \approx 41.15 \] Since \( x \) typically represents whole years in this context, we can round \( x \) to the nearest whole number: \[ x \approx 41 \] ### Conclusion: **The value of \( x \) when the average cost per patient per day reaches \$4,000 is approximately 41.** \[ x = \boxed{41} \]