Question 481375
let x = number of acres of strawberries
let y = number of acres of tomatoes
your limitation equations are:
land:
x + y <= 50
labor:
10x + 3y <= 300
fertilizer
8x + 20y <= 800
shipping space
x <= 26
y <= 37
solve all these equations for y so you can graph them.
you get:
land:
y <= 50-x
labor:
3y <= 300-10x becomes:
y <= 100 - 10x/3
fertilizer
20y <= 800-8x becomes:
y <= 40-8x/10 which becomes:
y <= 40 - 4x/5
shipping space equations stay the same at:
x <= 26
y <= 37
there are 2 other restrictions that are there, but not shown explicitly.
those are:
x >= 0
y >= 0
your graph would look like this:
{{{graph(600,600,-10,30,-10,40,50-x,100-(10x/3),40-(4x/10),100(x-26),37)}}}
a more distant view would look like this:
{{{graph(600,600,-10,60,-10,110,50-x,100-(10x/3),40-(4x/10),100(x-26),37)}}}
linear programming theory states that the maximum / minimum points would be at the intersection points and never in between.
since all of  your constraints have to be less than any line on the graph other than the x-axis and the y-axis, then your region of compatibility will be from x = 0 and y = 0 to any of the 7 intersection points.
those intersection points are:
<pre>
         x        y          profit
         0        0          0
         0        37         11,100
         7.5      37         14,100
         16.67    33.33      16,666.67
         21.43    28.57      17,142.86
         26       0          10,400
         27       13.33      14,400
</pre>
the profit equation is P = 400x + 300y.
you solve the profit equation at each of the intersection points and then choose the intersection point that provides you with the maximum profit.
the maximum profit point is when x = 21.43 and y = 28.57
that would be 21.43 acres of strawberries and 28.57 acres of tomatoes.
here's a reference that explains the process.
<a href = "http://www.purplemath.com/modules/linprog.htm" target = "_blank">http://www.purplemath.com/modules/linprog.htm</a>