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Question 617649: The available parking area of a parking lot is 900 square meters. A car requires 6 square meters of space and a bus requires 30 square meters of space. The attendant can handle no more than 70 vehicles.
Let c represent the number of cars, and b represent in the number of buses. Write system of inequalities to represent the number of cars and buses that can be parked on the lot.
Draw the graph showing the feasible region.
If the parking fees are $35.00 for cars and $30.00 for buses, how many of each type of vehicle should attendant accept to maximize income? What is the maximum income?
Parking fees for special events are $27.00 for cars and $25.00 for buses. How many of each vehicle should the attendant accept during a special event to maximize income? Whats the maximum income?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
If you have 900 square meters of space and each bus takes 30 square meters, then the maximum number of busses you can park in the space is 30. Hence:
If you have 900 square meters of space and each car takes 6 square meters, then the maximum number of busses you can park in the space is 150. Hence:
If the maximum number of vehicles the attendant can handle is 70,
Since you can't have a negative number of vehicles of either type:
and
On a set of coordinate axes labeled  and  (doesn't really matter which axis is which) graph all of the above inequalities. The solution set to the system of inequalities will be a quadrilateral in the first quadrant and this is your area of feasibility. HINT: Ordinarily when you graph an inequality, you shade in the half-plane that represents the solution set. However, when I graph a feasibiltiy area, I generally graph the inequalities with the OPPOSITE sense. The end result is that the feasibility area is totally unshaded and therefore much easier to differentiate from those parts of the plane that are not in the feasibility area.
I do complete linear programming optimization problems, complete with graphs and step by step explanations for $5 to $10 per problem depending on complexity. Write back and attach your problems for a quote.
Since you make $35 per day from each car and $30 per day from each bus, your revenue (and the objective function) is:
\ =\ 35c\ +\ 30b)
The optimum point to maximize your objective function is either one of the vertices of the feasibility polygon or in the event that two vertices give the same objective value, any point on that edge of the polygon is optimum.
Find the coordinates of each of the vertices of the feasibility polygon and evaluate the objective function for those values. Find the point that gives you the largest value for the objective function.
Repeat the process with the revised objective function:
\ =\ 27c\ +\ 25b)
Are you sure you have the prices for cars and busses in the right order? The way you have it, the outcome is not very interesting. It also doesn't make much sense that you would charge less to park a vehicle that takes up 5 times the space.
I do complete linear programming optimization problems, complete with graphs and step by step explanations for $5 to $10 per problem depending on complexity. Write back and attach your problems for a quote.
John
My calculator said it, I believe it, that settles it
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