SOLUTION: A company uses buses and minibuses to transport a minimum of 800 and a
maximum of 1200 passengers per day. The number of passengers, x, transported
by buses must be a minimum of
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-> SOLUTION: A company uses buses and minibuses to transport a minimum of 800 and a
maximum of 1200 passengers per day. The number of passengers, x, transported
by buses must be a minimum of
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Question 1141155: A company uses buses and minibuses to transport a minimum of 800 and a
maximum of 1200 passengers per day. The number of passengers, x, transported
by buses must be a minimum of 400 per day, but may not exceed more
than three times the number of passengers, y, transported by minibuses per
day. (Note that x, y 2 N.)
1 Represent the above constraints by a system of inequalities and indicate
the feasible region graphically.
2 The profit per day per passenger travelling by bus is R1, 50 and the
profit per day per passenger travelling by minibus is R1, 20. Determine
from your graph the values of x and y for which the profit will be a
maximum. Answer by Theo(13342) (Show Source):
those equations represent the fact that a minimum of 800 people and a maximum of 1200 people can be transported by either bus or minibus.
x >= 400
x <= 3 * y
those equations represent the fact that no less than 400 passengers and no more than 3 times the number of passengers transported by minibus can be transported by bus.
x >= 0
y >= 0
those inequalities represent the fact that the number of passengers can't be negative.
the profit per day by bus is 50 and the profit per day by minibus.
that's your objective function that you want to maximize.
your objective function is profit = 50 * x + 20 * y.
using the demos.com calculator, you will graph the opposite of your constraints and then evaluate your objective at each of the corner points of your feasible region.
you feasible region will be the area on the graph that is NOT shaded.
this can be done with the desmos.com calculator, making it a lot easier to see the region of feasibility.
your constraint functions are:
x + y >= 800
x + y <= 1200
x >= 400
x <= 3 * y
x >= 0
y >= 0
note that the desmos.com calculator also allows you to graph these functions without solving them for y first.
both these facts are huge improvements over other graphing calculator that do not have this capability.
your graph will look like this:
that's the easy way.
all the constraint have been satisfied.
the maximum profit is when 900 passengers are transported by bus and 300 passengers are transported by minibus.
the way you are probably most familiar with, or would be most familiar with if you didn't know about the desmos.com calculator is as follows:
your constraint functions are, once again.
x + y >= 800
x + y <= 1200
x >= 400
x <= 3 * y
x >= 0
y >= 0
you would graph the equality portions of these equations and then you would shade the area of the graph that satisfies the inequality portion of those equations.
you would also solve these inequalities for y first.
x + y >= 800 becomes y >= 800 - x and you would graph y = 800 - x.
x + y <= 1200 becomes y <= 1200 - x and you would graph y = 1200 - x.
x >= 400 becomes x >= 400 and you would graph x = 400.
x <= 3 * y becomes y >= x/3 and you would graph y = x/3.
x >= 0 becomes x >= 0 and you would graph x = 0.
y >= 0 becomes y >= 0 and you would graph y = 0
before you shade the regions of the graph, it would look like this.
after you shade the regions of the graph, it would look like this.
