Question 1171437
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The constraints are....<br>
(1) {{{x+3y<=60}}}
(2) {{{x+y>=10}}}
(3) {{{x-y<=0}}}
(4) {{{x>=0}}}
(5) {{{x>=0}}}<br>
Undoubtedly the last is supposed to be {{{y>=0}}}; however, with the other constraints, that one is unnecessary.<br>
In this particular problem, finding the maximum value of the objective function is easy, using constraint (1):<br>
{{{P = 5x+15y = 5(x+3y) <= 5(60) = 300}}}<br>
So the maximum value of the objective function is going to be 300.<br>
In the linear programming solution to the problem using a graph, this will correspond to the maximum value of the objective function being obtained anywhere along the portion of the graph of {{{x+3y=60}}} that satisfies the other constraints.<br>
The corners of the feasibility region are...<br>
A(0,10)  (intersection of (2) and (4)
B(0,20)  (intersection of (1) and (4)
C(5,5)  (intersection of (2) and (3)
D(15,15)  (intersection of (1) and (3)<br>
The objective function evaluated at those points is<br>
A: 15(10) = 150
B: 15(20) = 300
C: 5(5)+15(5) = 25+75 = 100
D: 5(15)+15(15) = 75+225 = 300<br>
Subject to the given constraints,<br>
The minimum value of the objective function is 100, at (5,5).
The maximum value of the objective function is anywhere on the line x+3y=60 between (0,20) and (15,15).<br>