Question 304513
The maximum number of ticket you can sell are 36 children and 28 adult = 64 tickets.


The maximum revenue you can make is 28 * 7 plus 36 * 4 = $340.00.


The minimum revenue you can make is $112.


Let x = number of adult tickets sold.


Let y = number of children tickets sold.


Your constraint equations are:

x >= 0
y >= 0
x <= 28
y <= 36
7x + 4y <= 340 (max revenue)
7x + 4y >= 112 (min revenue)


Your revenue equation is:


7*x + 4*y = R


You would graph all 6 constraint equations.


To graph the last 2 equations, you would solve for y.


They become:


y <= (340-7x)/4 (max revenue)
y >= (112-7x)/4 (min revenue)


Your graph would look like this:


{{{graph(600,600,-100,100,-100,100,(340-7x)/4,(112-7x)/4,36)}}}


The line that crosses the y-axis at 85 is your maximum revenue line.


The line that crosses the y-axis at 28 is your minimum revenue line.


The line that crosses the y-axis at 36 is your maximum number of children tickets line.


The x-axis is your minimum number of children tickets line.


The maximum number of adult tickets line at x = 28 can't be drawn on this graph, but can be shown on the picture below. 


The y-axis is your minimum numbe3r of adult tickets line.  


Your answer must lie in the following region.


It must be on or to the right of the vertical line at x = 0 (minimum adult tickets line).
It must be on or to the left of the vertical line at x = 28 (maximum adult tickets line).
It must be on or above the horizontal line at y = 0 (minimum children tickets line).
It must be on or below the horizontal line at y = 36 maximum children tickets line).
It must be on or below and to the left of the maximum revenue line.
It must be on or above and to the right of the minimum revenue line.


A picture of your graph with the feasible region is shown below:


<font color = "red"><img src = "http://theo.x10hosting.com/problems/304513.jpg" alt = "***** picture not found *****" /></font>


The graph does not show the revenue.


That has to be calculate from the information on the graph.


The Revenue Equation is, once again, R = 7*x + 4*y.


Looking at the Minimum Revenue Line:


When x = 0 and y = 28, the revenue is equal to 7*0 + 4*28 = $112.00 which is the minimum revenue that can be made.


When x = 16 and y = 0, the revenue is equal to 7*16 + 4*0 = $112.00 which is the minimum revenue that can be made.


Any point on the minimum revenue line in between the x-axis and the y-axis is a mix of adult and children tickets that will total up to $112.00.


Looking at the maximum revenue line:


When x = 0 and y = 85, the revenue is equal to 7*0 + 4*85 = $340.00 which is the maximum revenue that can be made.


That maximum revenue, however, cannot be attained when x = 0 because y cannot be greater than 36.   As you can see from the picture, y = 85 is not in the feasible region.


When y = 0 and x = 48.57142857, the revenue is equal to 48.l57142857*7 + 0*4 = $340.00 which is the maximum revenue that can be made.


That maximum revenue, however, cannot be attained when y = 0 because x cannot be greater than 28.   As you can see, x = 48.157142857 is not in the feasible region.


The intersection of the lines y = 36 and x = 28 and the maximum revenue line are, however, in the feasible region.


When x = 28 and y = 36, the revenue is 7*28 + 4*36 = $340 which is the maximum revenue that can be attained while still being within the feasible region.


By looking at the intersection points of all the constraint lines on the graph, you will find the maximum or the minimum revenue equation.


It was at the intersection of the maximum revenue line and the maximum number of adult tickets line and the maximum number of children tickets line that we found the maximum revenue that could be attained and still be within the feasible region.


It was at the intersection of the minimum revenue line and the maximum number of children tickets line and the minimum number of adult tickets line that we found the minimum revenue that could be attained and still be within the feasible region.


It was also at the intersection of the minimum revenue line and the minimum number of children tickets line and the maximum number of adult tickets line that we found the minimum revenue that could be attained and still be within the feasible region.


Your question was to find the feasible region on the graph..


It's the shaded region in the picture of the graph.


Any value of x and y that lies within that region is feasible.


This means that the revenue will be greater than or equal to $112.00 and less than or equal to $340.


For example:


x = 20 and y = 20 lies within that region.


Revenue would be 7*20 + 4*20 = $220.00


The region is feasible because all constraints have been met.


When x = 5 and y = 5 the revenue is equal to 7*5 + 4*5 = $55.00


While the number of tickets are within the constraints, the revenue is not within the constraints because the revenue is less than $112.00.


The point (5,5) is not within the feasible region as can be seen in the picture of the graph.