Find the maximum and minimum values of f(x,y)=2y+x for the polygonal convex set determined by this system?

Question

of inequalities: x > or equal to 1; y > or equal to 0; x < or equal to 4-y

a. minimum: 1; maximum: 4
b. minimum: 0; maximum: 5
c. minimum: 1; maximum: 7
d. minimum: 4; maximum: 8

Answer 1

Draw an xy-plane, and on it draw:

== the verttical line x = 1 ==> you want the points to its right, since x >= 1;

== The horizontal line y = 0 (this is just the x-axis) ==> you want the points above this line, sin ce y >= 0;

== the line y = -x + 4 <==> x = 4-y ==> you want the points below this line since x <= 4 - y.

You're left with a straight triangle with vertices at (1,0), (1,3) and (4,0). A well-known theorem in lineal planning tells us that the target function f(x,y) = 2y + x will attain its maximum and minimum values on this polygon's vertices:
F(1,0) = 1
F(1,3) = 7
F(4,0) = 4
So the minimum value is 1, and the maximum one is 7 ==> option (c)
Regards
Tonio

Answer 2

c. minimum: 1; maximum: 7

Answer 3

? for this reason area is proscribed as a million <= x <= 3; besides x/2 +y <= 5, for this reason y <= 5 -x/2; for this reason variety for y is y1 <= y <=y2, the place y1=y(3) =5 –3/2, y2= y(a million) =5 –a million/2, so 3.5 <= y <= 4.5; ? min f(x) =2*x[min] –y[max] +2 =2*a million –4.5 +2 =-0.5;