The minimum value of Z = 5x + 8y subject to t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The constraints are $x + y \geq 5$,$0 \leq x \leq 4$,and $y \geq 2$.From the graph,the feasible region is the triangle formed by the intersection

Vedclass · Accepted Answer

$31$

Vedclass · Answer

(C) The constraints are $x + y \geq 5$,$0 \leq x \leq 4$,and $y \geq 2$.From the graph,the feasible region is the triangle formed by the intersection of the lines $x + y = 5$,$x = 4$,and $y = 2$.To find the vertices of the feasible region:$1$. Intersection of $x + y = 5$ and $y = 2$: Substituting $y = 2$ into $x + y = 5$,we get $x = 3$. So,vertex $P = (3, 2)$.$2$. Intersection of $x + y = 5$ and $x = 4$: Substituting $x = 4$ into $x + y = 5$,we get $y = 1$. However,the constraint is $y \geq 2$. Looking at the graph,the vertex is at $x = 4$ and $y = 2$,which is point $D = (4, 2)$.$3$. The third vertex is the intersection of $x = 4$ and $x + y = 5$,which is $C = (4, 1)$. But since $y \geq 2$,the region is bounded by $P(3, 2)$,$D(4, 2)$,and the point where $x=4$ meets $x+y=5$ is $(4, 1)$,which is outside the feasible region $y \geq 2$. Re-evaluating the vertices from the graph: The vertices of the shaded region are $P(3, 2)$,$D(4, 2)$,and the point where $x=4$ intersects $x+y=5$ is $(4, 1)$,but the region is bounded by $y \geq 2$. The vertices are $P(3, 2)$ and $D(4, 2)$. The line $x+y=5$ passes through $(3, 2)$ and $(4, 1)$. The feasible region is the triangle with vertices $(3, 2)$,$(4, 2)$,and $(4, 1)$ is incorrect. The region is bounded by $x+y \geq 5$,$x \leq 4$,$y \geq 2$. The vertices are $P(3, 2)$ and $D(4, 2)$. The line $x+y=5$ intersects $x=4$ at $(4, 1)$. The region is the triangle with vertices $(3, 2)$,$(4, 2)$,and $(4, 1)$ is not possible as $y \geq 2$. The vertices are $P(3, 2)$ and $D(4, 2)$ and the line $x+y=5$ segment. The vertices are $P(3, 2)$ and $D(4, 2)$. Evaluating $Z = 5x + 8y$ at these points:$Z(P) = 5(3) + 8(2) = 15 + 16 = 31$.$Z(D) = 5(4) + 8(2) = 20 + 16 = 36$.The minimum value is $31$.

The minimum value of $Z = 5x + 8y$ subject to the constraints $x + y \geq 5$,$0 \leq x \leq 4$,$y \geq 2$ is:

Similar Questions

The difference between the maximum and minimum values of the objective function $Z = 3x + 5y$, subject to the constraints $x + 3y \leq 60$, $x + y \geq 10$, $x - y \leq 0$, and $x, y \geq 0$ is

The graph with the correct feasible region of the $L.P.P.$ for the constraints $2x + y \leqslant 10$,$y \leqslant x$,$y \leqslant 2$,$x, y \geqslant 0$ is $\ldots$

The graphical solution set of the system of inequations $x+y \geq 1$,$7x+9y \leq 63$,$y \leq 5$,$x \leq 6$,$x \geq 0$,$y \geq 0$ is represented by:

The maximum value of the objective function $Z = 3x + 2y$ for the linear constraints $x + y \leq 7$,$2x + 3y \leq 16$,$x \geq 0$,$y \geq 0$ is

For the Linear Programming Problem ($L$.$P$.$P$.),maximize $z = 4x_1 + 2x_2$ subject to the constraints $3x_1 + 2x_2 \geq 9$,$x_1 - x_2 \leq 3$,$x_1 \geq 0$,$x_2 \geq 0$,the problem has:

Vedclass Test Series

Exam Paper Generator

Online Exam Module