Corner points of the bounded feasible region | vedclass

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Question

(B) To find the minimum and maximum values of the objective function $z = 8x + 12y$,we evaluate $z$ at each corner point:$1. 	ext{At } (0,4): z = 8(0

Vedclass · Accepted Answer

$(i) (0,4), (ii) (12,16), (iii) 288, (iv) 48$

Vedclass · Answer

(B) To find the minimum and maximum values of the objective function $z = 8x + 12y$,we evaluate $z$ at each corner point:$1. 	ext{At } (0,4): z = 8(0) + 12(4) = 48$$2. 	ext{At } (6,0): z = 8(6) + 12(0) = 48$$3. 	ext{At } (12,0): z = 8(12) + 12(0) = 96$$4. 	ext{At } (12,16): z = 8(12) + 12(16) = 96 + 192 = 288$$5. 	ext{At } (0,10): z = 8(0) + 12(10) = 120$Comparing these values:- The minimum value is $48$,which occurs at both $(0,4)$ and $(6,0)$.- The maximum value is $288$,which occurs at $(12,16)$.Matching the given requirements:$(i)$ Minimum value occurs at $(0,4)$ (or $(6,0)$).$(ii)$ Maximum value occurs at $(12,16)$.$(iii)$ Maximum of $z$ is $288$.$(iv)$ Minimum of $z$ is $48$.Thus,option $(B)$ is the correct match.

Column	Value
$A$. Maximum of $z$	$300$
$B$. Constant value	$325$

Similar Questions

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:

The corner points of the feasible region determined by the system of linear constraints are $(0,0), (0,40), (20,40), (60,20), (60,0)$. The objective function is $z=4x+3y$. Compare the quantity in Column $A$ and Column $B$.

Column Value

$A$. Maximum of $z$ $300$

$B$. Constant value $325$

Minimise $Z = 3x + 2y$ subject to the constraints:
$x + y \geqslant 8$ ... $(1)$
$3x + 5y \leqslant 15$ ... $(2)$
$x \geqslant 0, y \geqslant 0$ ... $(3)$

Solve the Linear Programming Problem graphically:
Minimise $Z = 3x + 5y$
subject to the constraints:
$x + 3y \geq 3$
$x + y \geq 2$
$x, y \geq 0$

The minimum value of $Z = 2x + 3y$ for the system of linear constraints: $2x + 4y \leq 12$,$x + y \leq 3$,$x \geq 0$,and $y \geq 0$ is . . . . . . .

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