Question

alonzo does a weekly exercise program consisting of cardiovascular work and weight training. each week, he exercises for at least 12 hours. he spends at most 9 hours doing cardiovascular work. he spends at most 10 hours on weight training. let x denote the time (in hours) that alonzo spends doing cardiovascular work. let y denote the time (in hours) that he spends on weight training. shade the region corresponding to all values of x and y that satisfy these requirements.

Explanation:

Step1: Identify Constraints

• Total exercise: \( x + y \geq 12 \) (at least 12 hours).
• Cardiovascular: \( 0 \leq x \leq 9 \) (non - negative, at most 9 hours).
• Weight training: \( 0 \leq y \leq 10 \) (non - negative, at most 10 hours).

Step2: Graph Each Constraint

• For \( x + y \geq 12 \): The line \( x + y=12 \) has intercepts \((12,0)\) and \((0,12)\). Shade above the line (since \( x + y\geq12\)).
• For \( 0\leq x\leq9 \): Shade between \( x = 0 \) and \( x=9 \) (vertical lines).
• For \( 0\leq y\leq10 \): Shade between \( y = 0 \) and \( y = 10 \) (horizontal lines).

Step3: Find Intersection Region

The feasible region is the intersection of all shaded areas. It is bounded by:

• \( x = 0 \), \( x = 9 \), \( y = 0 \), \( y = 10 \), and \( x + y=12 \).
• The intersection of \( x + y = 12 \) with \( x = 9 \) gives \( y=3 \) (point \((9,3)\)).
• The intersection of \( x + y = 12 \) with \( y = 10 \) gives \( x = 2 \) (point \((2,10)\)).

So the feasible region is a polygon with vertices at \((2,10)\), \((9,10)\), \((9,3)\), \((2,10)\) (wait, actually, let's correct: vertices are \((2,10)\) (from \(x + y=12\) and \(y = 10\)), \((9,3)\) (from \(x + y=12\) and \(x = 9\)), \((9,0)\) (but \(y\geq0\), but also \(x + y\geq12\), so actually the lower - bound is from the intersection of the constraints. Wait, more accurately, the feasible region is where \(2\leq x\leq9\), \(3\leq y\leq10\) (from \(x + y\geq12\), \(x\leq9\) implies \(y\geq12 - 9 = 3\); \(y\leq10\) implies \(x\geq12 - 10 = 2\)), and \(x\geq0\), \(y\geq0\), \(x\leq9\), \(y\leq10\). So the region to shade is the area where \(x\) is between 2 and 9, \(y\) is between 3 and 10, and also satisfying \(x + y\geq12\), \(x\leq9\), \(y\leq10\), \(x\geq0\), \(y\geq0\).

To shade:

1. Draw the line \(x + y=12\) (dashed or solid? Since the inequality is \(\geq\), it's solid).
2. Shade above \(x + y = 12\), between \(x = 0\) and \(x = 9\), and between \(y = 0\) and \(y = 10\). The overlapping region is the feasible region.

(Note: Since this is a graph - shading problem, the key is to identify the correct boundaries as above and shade the region that satisfies all the inequalities.)

Answer:

The feasible region is the area bounded by \(x = 2\) to \(x = 9\), \(y = 3\) to \(y = 10\), and the line \(x + y=12\), within the rectangles \(0\leq x\leq9\) and \(0\leq y\leq10\). To shade, follow the steps of graphing each constraint and finding their intersection.