Question 1102103
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<pre>
First inequality 9x + 12y >= 61 is THIS restriction

     y >= {{{(61-9x)/12}}}.


Second inequality 8x + 4y >= 32 is THIS restriction

    y >= {{{(32-8x)/4}}}.


Together with the inequalities x >= 0, y >= 0 they form THIS feasibility domain in the first quadrant QI, shown in the Figure below:


{{{graph( 330, 330, -2.5, 12.5, -2.5, 12.5,
          (61-9x)/12,  (32-8x)/4
)}}}


Plot y >= {{{(61-9x)/12}}} (over the read line) and  y >= {{{(32-8x)/4}}} (over the green line)

Feasibility domain is <U>INFINITE AREA</U> in Q1 <U>OVER the both red and green lines</U>.



Feasibility domain has 3 vertices:

    P1 = (0,8)        (y-intercept to green line);

    P2 = (7/3,10/3)   (intersection point of the red and green line);

    P3 = (61/9,0)     (x-intercept to red line).


You should calculate the value of the objective function Z = 5x + 30y  at these three points:

    at  P1:   Z = 5*0      + 30*8       = 240;

    at  P2:   Z = 5*(7/3)  + 30*(10/3)  = 111.667;

    at  P3:   Z = 5*(61/9) + 30*0       = 33.889.


The minimum is achieved at the point P3 = (61/9,0), where x= 61/9, y=0,  and is equal to 305/9 = 33.889 (approximately).
</pre>

To see other mini-max problems solved by the Linear Programming method, look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Solving-minimax-problems-by--the-Linear-Programming-method.lesson>Solving minimax problems by the Linear Programming method</A> 

in this site.