\( 1 \leftarrow \) The average cost per hour in dollars of producing \( \times \) nding lawn mowers is given by the following \( \overline{\mathrm{C}}(\mathrm{x})=0.8 \mathrm{x}^{2}+30 \mathrm{x}-270+\frac{2900}{\mathrm{x}} \) (a) Use a graphing utility to delermine the number of riding lawn mowers to produce in order to minimize average cost (b) What is the minimum average cost? (a) The average cost is minimized when approxmately \( \square \) lawn mowers are produced per hour (Rcund to the nearest whole number as needed)

1. We begin with the average cost function \[ \overline{C}(x)=0.8x^2+30x-270+\frac{2900}{x}. \] 2. To find the production level that minimizes the cost, we first compute the derivative. Differentiating term‐by‐term gives \[ \overline{C}'(x)=1.6x+30-\frac{2900}{x^2}. \] 3. Setting \(\overline{C}'(x)=0\) to locate the minimum, we have \[ 1.6x+30-\frac{2900}{x^2}=0. \] 4. Rearranging, \[ 1.6x+30=\frac{2900}{x^2}. \] 5. Multiplying both sides by \(x^2\) (with \(x>0\)) yields \[ 1.6x^3+30x^2-2900=0. \] 6. Solving the cubic equation numerically, one finds \[ x\approx8.2. \] Rounding to the nearest whole number, we produce approximately \(8\) riding lawn mowers per hour. 7. To determine the minimum average cost, substitute \(x\approx8.2\) back into the cost function: \[ \overline{C}(8.2)=0.8(8.2)^2+30(8.2)-270+\frac{2900}{8.2}. \] First, compute \[ 8.2^2=67.24, \] so \[ 0.8(67.24)=53.792, \] and \[ 30(8.2)=246. \] Now, \[ \frac{2900}{8.2}\approx353.66. \] Combining these values: \[ 53.792+246-270+353.66\approx383.45. \] 8. Therefore, the minimum average cost is approximately \(\$383.45\) per hour. Final Answers: (a) Approximately \(8\) lawn mowers should be produced per hour. (b) The minimum average cost is about \(\$383.45\) per hour.