Question

= 6x + 2y Find the maximum and minimum values (if they exist) of : P ubject to the constraints of: 4x + y 2 12 x + y s 14 x - 3y s6 x 20; y 20

Answer 1

the first constraint is

$4x+y\ge 12$

divide by 12

$\frac{4x}{12}+\frac{y}{12}\ge \frac{12}{12}$

$\frac{x}{3}+\frac{y}{12}\ge 1$

compare with the line $\frac{x}{a}+\frac{y}{b}= 1$

so here x-intercept is (a,0)=(3,0)

and the y-intercept is (0,b)=(0,12)

draw a line between (3,0) and (0,12)

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the second constraint is

$x+y\le 14$

divide by 14

$\frac{x}{14}+\frac{y}{14}\le \frac{14}{14}$

$\frac{x}{14}+\frac{y}{14}\le 1$

compare with the line $\frac{x}{a}+\frac{y}{b}= 1$

so here x-intercept is (a,0)=(14,0)

and the y-intercept is (0,b)=(0,14)

draw a line between (14,0) and (0,14)

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the third constraint is

$x-3y\le 6$

$\frac{x}{6}-\frac{3y}{6}\le \frac{6}{6}$

$\frac{x}{6}-\frac{y}{2}\le 1$

$\frac{x}{6}+\frac{y}{-2}\le 1$

compare with the line $\frac{x}{a}+\frac{y}{b}= 1$

so here x-intercept is (a,0)=(6,0)

and the y-intercept is (0,b)=(0,-2)

draw a line between (6,0) and (0,-2)

(0,14) 14 12 (0,12) 10 6 (12, 2) 2 (3, 0) (6,0) 2 10 2 6 12 14

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The value of the objective function at each of these extreme points is as follows:

Extreme Point Coordinates (x,y)	Objective function value z=6x+2y
A(0,12)	6(0)+2(12)=24
B(0,14)	6(0)+2(14)=28
C(12,2)	6(12)+2(2)=76
D(6,0)	6(6)+2(0)=36
E(3,0)	6(3)+2(0)=18

The maximum value of the objective function z=76 occurs at the extreme point (12,2).

Hence, the optimal solution to the given LP problem is x=12,y=2 and max z=76.

${\color{Red} (x,y)=(12,2), \:\:\:\:\:\:\:Max\:Z=76}$

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The minimum value of the objective function z=18 occurs at the extreme point (3,0).

Hence, the optimal solution to the given LP problem is x=3,y=0 and min z=18.

${\color{Red} (x,y)=(3,0), \:\:\:\:\:\:\:Min\:Z=18}$