They are:
suites and cabins.
Let x = number of suites.
Let y = number of cabins.
Each suite can accommodate 6 passengers.
Each cabin can accommodate 3 passengers.
Maximum number of passengers is 330.
Equation for passengers is:
6x + 3y <= 330
Each suite occupies 50 square meters of floor space.
Each cabin occupies 10 square meters of floor space.
Maximum amount floor space is 2300 square meters.
Equation for floor space is
50x + 10y <= 2300
Weekly income for suites is $1000.
Weekly income for each cabin is $300.
Equation for income is:
I = 1000*x + 300*y
You want to maximize the income.
You want to determine what that maximum is.
You know that x has to be >= 0 and you know that y has to be >= 0.
Your 4 constraint equations are
6x + 3y <= 330
50x + 10y <= 2300
x >= 0
y >= 0
You need to graph these to find out what the intersection points are.
Solve for y in each equation to get:
y <= (330-6x)/3
y <= (2300-50x)/10
Graph the equations of:
y = (330-6x)/3
y = (2300-50x)/10
The graph of these equations looks like this:
The graph shows that there are 5 intersection points.
2 of those intersections occur when x = 0.
2 of those intersections occur when y = 0.
1 of those intersections occur when x and y are not equal to 0.
You can find the 0 intersections points easily enough.
Your 2 constraint equations are:
y <= (330-6x)/3
y <= (2300-50x)/10
When x = 0, these equations become:
y <= 330/3
y <= (2300)/10
This makes:
y <= 110
y <= 230
Since y has to be smaller than equal to both, then y has to be smaller than or equal to 110 when x = 0.
When y = 0, these equations become:
0 <= (330-6x)/3
0 <= (2300-50x)/10
Solve for x in both equations to get:
x <= 55
x <= 46
Since x has to be smaller than 55 and smaller than 46, then x has to be smaller than or equal to 46.
When x = 0, y has to be smaller than or equal to 110.
When y = 0, x has to be smaller than or equal to 46.
When x and y are not equal to 0, you need to solve the equations simultaneously to come up with the intersection point.
Set the original equations up as an equality rather than an inequality.
You get:
6x + 3y = 330
50x + 10y = 2300
Solve for the intersection of these lines to get an x-value and a y-value that satisfies both equations simultaneously.
Multiply the first equation by 10 and the second equation by 3 to get:
60x + 30y = 3300
150x + 30y = 6900
Subtract the first equation from the second to get:
90x = 3600
Divide both sides of this equation by 90 to get:
x = 40
Substitute in both original equations to solve for y.
Original equations are:
6x + 3y = 330
50x + 10y = 2300
Substitute 40 for x to get:
6*40 + 3y = 330
50*40 + 10y = 2300
Simplify to get:
240 + 3y = 330
2000 + 10y = 2300
Subtract 240 from both sides of first equation and subtract 2000 from both sides of second equation to get:
3y = 330 - 240 = 90
10y = 2300 - 2000 = 300
Divide both sides of first equation by 3 and divide both sides of second equation by 10 to get
y = 30
y = 30
Since y = 30 in both equations, you're good.
The intersection point of these 2 lines is (x,y) = (40,30).
This intersection point is valid since the equations say <= and the intersection points of these equations are the = part of that.
The rules of solving linear equations involving maximum / minumum with constraint equations are that the maximum / minimum points will be at the intersections.
Our 3 intersection points that are valid are:
(x,y) = (0,110)
(x,y) = (46,0)
(x,y) = (40,30)
A picture of the graph above showing the inequality and the region of interest is shown below:
The region of interest is in brown.
That's the region bounded by (0,0), (0,110), (40,30), and (46,0).
(0,0) is a valid point but it generates no income so it was not considered in the evaluation of the income equation at the valid points of intersection.
We need to evaluate the income equation at the 3 valid points of (0,110), (40,30), and (46,0)
The income equation is:
I = 1000*x + 300*y
When x = 0 and y = 110, I = $33,000.
When y = 46 and y = 0, I = $46,000.
When x = 40 and y = 30, I = $49,000.
Maximum Income will occur when x = 40 and y = 30.
That maximum income will be $49,000.
Let's see if the constraints have been met.
When x = 0 and y = 110, number of passengers = 3*110 = 330 <= 330 (ok).
When x = 46 and y = 0, number of passengers = 6*46 = 276 <= 330 (ok).
When x = 40 and y = 30, number of passengers = 6*40 + 3*30 = 240 + 90 = 330 <= 330 (ok).
When x = 0 and y = 55, floor space = 10*55 = 550 <= 2300 (ok).
When x = 46 and y = 0, floor space = 50*46 = 2300 <= 2300 (ok).
When x = 40 and y = 30, floor space = 50*40 + 10*30 = 2000 + 300 = 2300 <= 2300 (ok).
All the constraints are met so the maximum income occurs when x = 40 and y = 30 and the maximum income is $49,000.
Any other combination not at the intersection points will yield an income that will not exceed $49,000.
For example, When x = 41, y has to be <= 28 in the first equation and y has to be <= 25 in the second equation. To satisfy both equations, y has to be <= 25.
In the income equation, 41*1000 + 300*25 = $48,500 <= $49,000.
You can try any other points not at the intersections of the constraint equations and you will see that the maximum income is at the intersections.