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Question
- as part of kaylas exercise program, she either runs 6 miles/day or rides her bike 10 miles/day. her new goal is to cover a minimum distance of 200 miles, with at least 15 of the days running. she would like to determine the number of days it would take to accomplish this. a) write a system of linear inequality to represent this situation, then graph. b) using your graph, give two possible combinations of days kayla can run and ride her bike to meet her goal. miles running and miles biking.
Step1: Define the variables
Let $x$ be the number of days running and $y$ be the number of days biking.
Step2: Write the first - inequality for distance
She runs 6 miles per day and bikes 10 miles per day, and her goal is to cover at least 200 miles. So, $6x + 10y\geq200$. Simplify it to $3x + 5y\geq100$.
Step3: Write the second - inequality for running days
She has at least 15 days running, so $x\geq15$.
Step4: Non - negativity constraints
Since the number of days cannot be negative, $x\geq0$ and $y\geq0$.
Step5: Graph the inequalities
For $3x + 5y\geq100$, find the intercepts. When $x = 0$, $y = 20$; when $y=0$, $x=\frac{100}{3}\approx33.33$. Draw the line $3x + 5y = 100$ and shade the region above it. For $x\geq15$, draw the vertical line $x = 15$ and shade the region to the right of it. Also, consider the non - negativity constraints $x\geq0$ and $y\geq0$ (shade the first quadrant).
Step6: Find possible combinations
One possible combination: Let $x = 15$. Substitute into $3x + 5y\geq100$, we get $3\times15+5y\geq100$, $45 + 5y\geq100$, $5y\geq55$, $y\geq11$. So one combination is $x = 15$ (15 days running) and $y = 11$ (11 days biking).
Another combination: Let $x = 20$. Substitute into $3x + 5y\geq100$, we get $3\times20+5y\geq100$, $60 + 5y\geq100$, $5y\geq40$, $y\geq8$. So another combination is $x = 20$ (20 days running) and $y = 8$ (8 days biking).
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a) The system of linear inequalities is
b) 15 days running and 11 days biking; 20 days running and 8 days biking.