Let x = the number of airplanes to make.
Let y = the number of helicopters to make.

>>...Can't make more than 50 airplanes...<<

So that says

x < 50

>>...no more than 30 helicopters...<<

y < 30

>>...Cost to make airplane - 60,000...<<

That says 

the cost to make the x airplanes is 60000x

>>...Cost to make helicopter - 90,000...<<

That says:

The cost to make y helicopters is 90000y

>>...Budget to spend - 3.6 million...<<

That says 

60000x + 90000 < 3600000

So the constraints are:

              x < 50
              y < 30
60000x + 90000y < 3600000
              x > 0
              y > 0

Those last two are to prevent the number of 
planes or helicopters from being negative. 

We can simplify 

 60000x + 90000 < 3600000

by dividing through by 30000, getting

        2x + 3y < 120

So the constraints are

      x < 50
      y < 30
2x + 3y < 120
      x > 0
      y > 0

>>...get 400,000 for each sirplane sold...<<
>>...Cost to make airplane - 60,000...<<

So the profit made on an airplane is 
400000 - 60000 = 340000

So the profit made on x airplanes is 340000x

>>...850,000 for each helicopter sold...<<
>>...Cost to make helicopter - 90,000...<<

So the profit made on a helicopter is 
850000 - 90000 = 760000

So the profit made on y helicopters is 760000x

So the total profit P for making x airplanes and
y helicopters is

P = 340000x + 760000y

So to state the problem correctly:

Maximize the objective function

P = 340000x + 760000y

subject to the constraints:

      x < 50
      y < 30
2x + 3y < 120
      x > 0
      y > 0

We draw the graphs of these equations, formed
by replacing the symbols of inequality by
equal signs:

      x = 50  (a vertical line 50 units right of the y-axis)
              (the feasible region is on or left of this line) 
      y = 30  (a horizontal line 30 units above the x-axis
              (the feasible region is on or below this line) 
2x + 3y = 120 (a slanted line with intercepts (60,0), (0,40)
              (the feasible region is on or between this line
               and the x- and y- axes.  
      x = 0   (the y-axis)
              (the feasible region is on or right of the y-axis 
      y = 0   (the x-axis)
              (the feasible region is on or above the x-axis) 
 


The feasible region is this polygon:



Find the corner points by substitution using the 
equations:



Now substitute each of those corner points in the
objective function and you'll find that the optimum
point is (15,30) for a profit of $27,900,000 when
they make 15 airplanes and 30 helicopters.

Edwin