(B) To find the maximum value of the objective function $Z = 4x + 3y$,we evaluate $Z$ at each corner point of the feasible region:$1$. At $(0,0): Z =

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(B) To find the maximum value of the objective function $Z = 4x + 3y$,we evaluate $Z$ at each corner point of the feasible region:$1$. At $(0,0): Z = 4(0) + 3(0) = 0$$2$. At $(25,5): Z = 4(25) + 3(5) = 100 + 15 = 115$$3$. At $(16,16): Z = 4(16) + 3(16) = 64 + 48 = 112$$4$. At $(5,24): Z = 4(5) + 3(24) = 20 + 72 = 92$Comparing these values,the maximum value is $115$,which occurs at the point $(25,5)$.

Class 12 Mathematics Linear Programming MCQ based Question

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For a Linear Programming Problem $(LPP)$,if the objective function is $Z = 4x + 3y$ and the corner points of the bounded feasible region are $(0,0), (25,5), (16,16),$ and $(5,24)$,then the maximum value of $Z$ occurs at the point . . . . . . .

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A
$(0,0)$
B
$(25,5)$
C
$(16,16)$
D
$(5,24)$

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For a Linear Programming Problem $(LPP)$,if the objective function is $Z = 4x + 3y$ and the corner points of the bounded feasible region are $(0,0), (25,5), (16,16),$ and $(5,24)$,then the maximum value of $Z$ occurs at the point . . . . . . .

Similar Questions

The following graph represents a feasible region. The minimum value of $z = 5x + 4y$ is $\ldots \ldots$

The objective function of a Linear Programming Problem $(LPP)$ is

The corner points of the feasible region determined by the system of linear constraints are $(2, 72)$,$(15, 20)$,and $(40, 15)$. Let $Z = 6x + 3y$ be the objective function. The minimum value of $Z$ occurs at:

For the $LP$ problem,"Maximize $z = x + 4y$ subject to $3x + 6y \leq 6$,$4x + 8y \geq 16$ and $x \geq 0, y \geq 0$."

Show that the minimum of $Z$ occurs at more than two points.Maximize $Z = x + y$,subject to $x - y \leq -1$,$-x + y \leq 0$,$x, y \geq 0$.

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Show that the minimum of $Z$ occurs at more than two points.
Maximize $Z = x + y$,subject to $x - y \leq -1$,$-x + y \leq 0$,$x, y \geq 0$.