\begin{tabular}{l|l} Deluxe coffee is to be mixed with regular coffee to & To attain the minimum cost, use \( \square \) pounds \\ make at least 52 pounds of a blended coffee. The \\ mixture must contain at least 9 pounds of deluxe \\ coffee. Deluxe coffee costs \( \$ 4 \) per pound and \\ regular coffee \( \$ 3 \) per pound. How many pounds of \\ each kind of coffee should be used to \\ minimize costs? \end{tabular}

To minimize the cost of the blended coffee while meeting the specified constraints, let \( x \) be the pounds of deluxe coffee and \( y \) be the pounds of regular coffee. The objective is to minimize the cost function \( C = 4x + 3y \). The constraints are: 1. \( x + y \geq 52 \) (to make at least 52 pounds of coffee) 2. \( x \geq 9 \) (to include at least 9 pounds of deluxe coffee) 3. \( x, y \geq 0 \) (non-negativity constraints) To solve this, first, express \( y \) in terms of \( x \) using constraint 1: \[ y \geq 52 - x. \] Now, we need to consider significant values of \( x \) to comply with the constraints. Start with \( x = 9 \) (the minimum for deluxe coffee): \[ y \geq 52 - 9 = 43. \] Thus, at \( x = 9 \), \( y \) must be 43 pounds. The cost would then be: \[ C = 4(9) + 3(43) = 36 + 129 = 165. \] Next, test \( x = 10 \): \[ y \geq 52 - 10 = 42. \] Cost: \[ C = 4(10) + 3(42) = 40 + 126 = 166. \] Continue this process until it’s no longer feasible to reduce costs: For \( x = 43 \) (making \( y = 9\)): \[ y = 52 - 43 = 9. \] Cost: \[ C = 4(43) + 3(9) = 172 + 27 = 199. \] At \( x = 52 \) (making \( y = 0\)): \[ y = 0, C = 4(52) + 3(0) = 208. \] After calculating the costs for various values, the optimum combination is found at \( x = 9 \) and \( y = 43 \), yielding the minimum cost of $165. Therefore, you should mix **9 pounds of deluxe coffee and 43 pounds of regular coffee** to minimize costs.