Question
Maximize Z = 9x + 3y
Subject to
$2\text{x}+3\text{y}\leq13$
$3\text{x}+\text{y}\leq5$
$\text{x},\text{y}\geq0$
Subject to
$2\text{x}+3\text{y}\leq13$
$3\text{x}+\text{y}\leq5$
$\text{x},\text{y}\geq0$
✓
Answer
Coverting the given inequation into equations, we get
2x + 3y = 13, 3x + y = 5 and x = 0, y = 0
Region represented by $2\text{x}+3\text{y}\leq13:$
The line meets coordinate axes at $\text{A}_1\Big(\frac{13}{2},0\Big)$ and $\text{B}_1\Big(0,\frac{13}{3}\Big)$ respectively.
Join these points to obtain the line 2x + 3y = 13, clearly, (0,0) satisfies the in eqation $2\text{x}+3\text{y}\leq13$, so, the region in xy-plane that contains origin represents the solution set of $2\text{x}+3\text{y}\leq13$.
Region represented by $3\text{x}+\text{y}\leq5:$
The line meets coordinate axes at $\text{A}_2\Big(\frac{5}{3},0\Big)$ and $B_2(0, 5)$ respectively.
Join these points to obtain the line 3x + y = 5, clearly, (0, 0) satisfies the in eqation $3\text{x}+\text{y}\leq5$, so, the region in xy-plane that contains origin represents the solution set of $3\text{x}+\text{y}\leq5$.
Region represented by $\text{x},\text{y}\geq0:$
It clearly represent first quadrant of xy-plane.
The common region to regions represented by above in equalities.
The coordinates of the corner points of the shaded region are $\text{O}(0,0),\text{A}\Big(\frac{5}{3},0\Big),\text{P}\Big(\frac{2}{7},\frac{29}{7}\Big),\text{B}_2\Big(0,\frac{13}{3}\Big)$.
The value of Z = 9x + 3y at
$\text{O}(0,0)=9(0)+3(0)=0$
$\text{A}_1\Big(\frac{5}{3},0\Big)=9\Big(\frac{5}{3}\Big)+3(0)=15$
$\text{P}\Big(\frac{2}{7},\frac{29}{7}\Big)=9\Big(\frac{2}{7}\Big)+3\Big(\frac{29}{7}\Big)=15$
$\text{B}_2\Big(0,\frac{13}{3}\Big)=9(0)+3\Big(\frac{13}{3}\Big)=13$
Clearly, Z is maximum at every point on the line joining $A_1$ and P, So
$\text{x}=\frac{2}{7}$ or $\frac{2}{7}$, $\text{y}=0$ or $\frac{29}{7}$
and maximum Z = 15.
2x + 3y = 13, 3x + y = 5 and x = 0, y = 0
Region represented by $2\text{x}+3\text{y}\leq13:$
The line meets coordinate axes at $\text{A}_1\Big(\frac{13}{2},0\Big)$ and $\text{B}_1\Big(0,\frac{13}{3}\Big)$ respectively.
Join these points to obtain the line 2x + 3y = 13, clearly, (0,0) satisfies the in eqation $2\text{x}+3\text{y}\leq13$, so, the region in xy-plane that contains origin represents the solution set of $2\text{x}+3\text{y}\leq13$.
Region represented by $3\text{x}+\text{y}\leq5:$
The line meets coordinate axes at $\text{A}_2\Big(\frac{5}{3},0\Big)$ and $B_2(0, 5)$ respectively.
Join these points to obtain the line 3x + y = 5, clearly, (0, 0) satisfies the in eqation $3\text{x}+\text{y}\leq5$, so, the region in xy-plane that contains origin represents the solution set of $3\text{x}+\text{y}\leq5$.
Region represented by $\text{x},\text{y}\geq0:$
It clearly represent first quadrant of xy-plane.
The common region to regions represented by above in equalities.
The coordinates of the corner points of the shaded region are $\text{O}(0,0),\text{A}\Big(\frac{5}{3},0\Big),\text{P}\Big(\frac{2}{7},\frac{29}{7}\Big),\text{B}_2\Big(0,\frac{13}{3}\Big)$.
The value of Z = 9x + 3y at
$\text{O}(0,0)=9(0)+3(0)=0$
$\text{A}_1\Big(\frac{5}{3},0\Big)=9\Big(\frac{5}{3}\Big)+3(0)=15$
$\text{P}\Big(\frac{2}{7},\frac{29}{7}\Big)=9\Big(\frac{2}{7}\Big)+3\Big(\frac{29}{7}\Big)=15$
$\text{B}_2\Big(0,\frac{13}{3}\Big)=9(0)+3\Big(\frac{13}{3}\Big)=13$
Clearly, Z is maximum at every point on the line joining $A_1$ and P, So
$\text{x}=\frac{2}{7}$ or $\frac{2}{7}$, $\text{y}=0$ or $\frac{29}{7}$
and maximum Z = 15.
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