Weeding and watering times The cucumbers require 20 minutes of weeding and watering time, while the carrots require 15 minutes of weeding and watering. Paul and Art have up to 2 hours each day that can be used for weeding and watering both the cucumbers and the carrots. Start-up costs: Paul and Art are able to spend up to \( \$ 25 \) on carrot seed and cucumber starts. (Starts are small cucumber plants.) The cost for the cucumber plants is \( \$ 5 \) per plant while the cost for the carrot seeds is \( \$ 2 \) per package. Write a system of inequalities for the two constraints that Paul and Art have. Be surs to define your variables. a. \( \$ \) Solve the system of inequalities to show possible combinations of cucumbers and carrots Paul and Art could feasibly produce.

Let \( x \) = number of cucumber plants \( y \) = number of packages of carrot seeds The two constraints are: 1. Weeding and watering time. Each cucumber requires 20 minutes and each carrot (package) requires 15 minutes. With 2 hours available (i.e. \(120\) minutes), we have \[ 20x + 15y \le 120. \] Dividing the entire inequality by 5 gives \[ 4x + 3y \le 24. \] 2. Start‐up cost. Cucumbers cost \(\$5\) each and carrots cost \(\$2\) per package, with a budget of \(\$25\): \[ 5x + 2y \le 25. \] Also, since negative amounts of plants or seeds are not allowed, we require \[ x \ge 0 \quad \text{and} \quad y \ge 0. \] --- To graph the feasible region defined by \[ \begin{cases} 4x + 3y \le 24, \\ 5x + 2y \le 25, \\ x \ge 0, \; y \ge 0, \end{cases} \] we first find the intercepts and intersection of the lines: 1. For \(4x + 3y = 24\): - \(x\)-intercept: set \(y=0\) \(\Rightarrow\) \(4x=24\) so \(x=6\). - \(y\)-intercept: set \(x=0\) \(\Rightarrow\) \(3y=24\) so \(y=8\). 2. For \(5x + 2y = 25\): - \(x\)-intercept: set \(y=0\) \(\Rightarrow\) \(5x=25\) so \(x=5\). - \(y\)-intercept: set \(x=0\) \(\Rightarrow\) \(2y=25\) so \(y=12.5\). 3. Find the intersection point of the two lines by solving \[ \begin{aligned} 4x + 3y &= 24, \\ 5x + 2y &= 25. \end{aligned} \] Solve the second equation for \(y\): \[ 2y = 25 - 5x \quad \Rightarrow \quad y = \frac{25 - 5x}{2}. \] Substitute into the first equation: \[ 4x + 3\left(\frac{25 - 5x}{2}\right) = 24. \] Multiply both sides by 2: \[ 8x + 3(25 - 5x) = 48. \] Simplify: \[ 8x + 75 - 15x = 48 \quad \Rightarrow \quad -7x = 48 - 75 = -27, \] so \[ x = \frac{27}{7} \approx 3.857. \] Substitute \(x = \frac{27}{7}\) back to find \(y\): \[ y = \frac{25 - 5\left(\frac{27}{7}\right)}{2} = \frac{25 - \frac{135}{7}}{2} = \frac{\frac{175 - 135}{7}}{2} = \frac{\frac{40}{7}}{2} = \frac{20}{7} \approx 2.857. \] Thus, the intersection point is \[ \left(\frac{27}{7}, \frac{20}{7}\right). \] --- The system of inequalities that Paul and Art have is: \[ \begin{aligned} 20x + 15y &\le 120 \quad \text{or equivalently} \quad 4x + 3y \le 24, \\ 5x + 2y &\le 25, \\ x &\ge 0, \\ y &\ge 0. \end{aligned} \] Any combination \((x, y)\) satisfying the above will be feasible. For example: - If \(x=0\), then from \(4(0)+3y \le 24\) we have \(y \le 8\) and from \(5(0)+2y \le 25\) we have \(y \le 12.5\). Thus, \(y\) can be at most 8. - If \(y=0\), then from \(5x \le 25\) we get \(x \le 5\) and from \(4x \le 24\) we get \(x \le 6\). Thus, \(x\) can be at most 5. The intersection of both constraints defines the feasible region, which is the region below both lines in the first quadrant.