Question 864643
Graphing the system of inequalities representing the constraints:
The constraints {{{0<=x<=5}}} and {{{0<=y<=8}}}
determine the rectangle whose sides are parts of the lines
{{{x=0}}} , {{{x=5}}} , {{{y=0}}} , and {{{y=8}}} in quadrant I.
{{{drawing(300,300,-1,9,-1,9,
grid(0),red(line(5,0,5,8)),red(line(0,8,5,8))
)}}}
The constraints {{{4x + 5y<=30}}} and {{{4x + 3y<=20}}}
restrict the area further to the space below the lines
{{{4x + 5y=30}}} and {{{4x + 3y=20}}} :
{{{drawing(300,300,-1,9,-1,9,
grid(0),red(line(5,0,5,8)),red(line(0,8,5,8)),
blue(line(-1,6.8,8.75,-1)),blue(line(-1,8,5.75,-1))
)}}}
We know that it has to be below both lines, because point (0,0), the origin,
is a solution to both inequalities, {{{4x + 5y<=30}}} and {{{4x + 3y<=20}}} .
It is easy to draw those lines if you locate their x- and y-intercepts.
For {{{4x + 5y=30}}} :
{{{x=0}}}-->{{{4*0 + 5y=30}}}-->{{{0 + 5y=30}}}-->{{{5y=30}}}-->{{{y=30/5}}}-->{{{y=6}}} and
{{{y=0}}}-->{{{4x + 5*0=30}}}-->{{{4x+0=30}}}-->{{{4x=30}}}-->{{{x=30/4}}}-->{{{x=7.5}}}
so those intercepts are (0,6) and (7.5,0).
For {{{4x + 3y=20}}} :
{{{x=0}}}-->{{{4*0 + 3y=20}}}-->{{{0 + 3y=20}}}-->{{{3y=20}}}-->{{{y=20/3}}}-->{{{y=6&2/3}}}
{{{y=0}}}-->{{{4x + 3*0=20}}}-->{{{4x+0=20}}}-->{{{4x=20}}}-->{{{x=20/4}}}-->{{{x=5}}}
so those intercepts are (0,20/3) and (5,0).
 
Finding the corners of the "graphed region:"
The graph tells us that points (0,0) , (0, 6) and (5,0) are three of the corners of the "graphed region."
The remaining corner is the intersection of {{{4x + 5y=30}}} and {{{4x + 3y=20}}} .
To find the coordinates of that point we solve {{{system(4x + 5y=30,4x + 3y=20)}}} .
We start by subtracting the second equation from the first to find {{{y}}} :
{{{system(4x + 5y=30,4x + 3y=20)}}}-->{{{system(4x + 5y=30,2y=10)}}}-->{{{system(4x + 5y=30,y=10/2)}}}-->{{{system(4x + 5y=30,highlight(y=5))}}} .
Next, we substitute the value found for {{{y}}} in either equation, and solve to find {{{x}}} :
{{{system(4x+5y=30,y=5)}}}-->{{{4x+5*5=30}}}-->{{{4x+25=30}}}-->{{{4x=30-25}}}-->{{{4x=5}}}-->{{{highlight(x=5/4)}}}<-->{{{highlight(x=1.25)}}} .
So the fourth and last corner is (1.25,5).
 
Finding the value of the objective function {{{z = 11x - 25y}}} at each corner:
At (0,0), {{{z = 11*0 - 25*0=highlight(0)}}}
At (0, 6),  {{{z = 11*0 - 25*6=0-150=highlight(-150)}}}
At (5,0),  {{{z = 11*5 - 25*0=55-0=highlight(55)}}}
At (1.25,5),  {{{z = 11*1.25 - 25*5=13.75-125=highlight(-111.25)}}}
 
Finding the maximum and its location:
Now that we have the value at the four corners of the region limited by the constraints,
we see that the function maximum is {{{highlight(55)}}} at (5,0).
 
NOTE: The maximum (or minimum) of a linear function of x and y in a region that is a polygon must happen either at one corner, or at one edge.
If the greatest value found had been found at two corners, the entire edge connecting those two corners would have been the location of the maximum.