Maximize parking lot income given space and vehicle constraints
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We are given a parking lot optimization problem. We need to determine how many cars and buses to accept to maximize income, subject to area and vehicle count constraints. Let's visualize the constraints on a coordinate system where x represents cars and y represents buses.
Let's define our variables: x for the number of cars and y for the number of buses. We have three constraints: the area constraint 6x plus 30y must be less than or equal to 600, the vehicle count constraint x plus y must be less than or equal to 60, and non-negativity constraints. Our objective is to maximize income, which equals 2.5x plus 7.5y.
In linear programming, the maximum value of the objective function occurs at one of the corner points of the feasible region. We've identified four corner points: the origin, the x-axis intercept of the vehicle constraint, the y-axis intercept of the area constraint, and the intersection point of both constraints. Let's calculate the income at each point.
Now we calculate the income at each corner point using our objective function: Income equals 2.5x plus 7.5y. At the origin, income is zero. At point (60, 0), income is 150 dollars. At point (0, 20), income is also 150 dollars. At the intersection point (45, 15), income is 225 dollars. This is our maximum income.
Our solution is to accept 45 cars and 15 buses for a maximum income of 225 dollars. But let's verify this satisfies all constraints. For area: 6 times 45 plus 30 times 15 equals 270 plus 450 equals 720 square meters. This exceeds our 600 square meter limit! We need to recheck our calculations.
We made an error in identifying the corner points. The intersection of our constraints must satisfy both equations simultaneously. Let's solve them correctly. From the vehicle constraint, we get x equals 60 minus y. Substituting this into the area constraint gives us 6 times (60 minus y) plus 30y equals 600. Simplifying, we get 360 minus 6y plus 30y equals 600, which leads to 24y equals 240, so y equals 10. Therefore, x equals 50.
Now let's recalculate the income at each corner point with our corrected intersection point. At the origin, income is zero. At point (60, 0), income is 150 dollars. At point (0, 20), income is also 150 dollars. At our corrected point (50, 10), income is 2.5 times 50 plus 7.5 times 10, which equals 125 plus 75, or 200 dollars. Therefore, the maximum income is 200 dollars, achieved by accepting 50 cars and 10 buses.