A fruit grower can use two types of fertilize | vedclass

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

Question

(A) Let the fruit grower use $x$ bags of brand $P$ and $y$ bags of brand $Q$.The objective function is to maximize nitrogen: $Z = 3x + 3.5y$.Constrain

Vedclass · Accepted Answer

$140$ bags of $P$ and $50$ bags of $Q$; Maximum nitrogen = $595 \, kg$

Vedclass · Answer

(A) Let the fruit grower use $x$ bags of brand $P$ and $y$ bags of brand $Q$.The objective function is to maximize nitrogen: $Z = 3x + 3.5y$.Constraints based on the table:$1$. Phosphoric acid: $x + 2y \geq 240$$2$. Potash: $3x + 1.5y \geq 270 \implies 2x + y \geq 180$$3$. Chlorine: $1.5x + 2y \leq 310$$4$. Non-negativity: $x, y \geq 0$Solving the system of inequalities,the feasible region is bounded by the corner points $A(140, 50)$,$B(20, 140)$,and $C(40, 100)$.Evaluating $Z = 3x + 3.5y$ at corner points:- At $A(140, 50): Z = 3(140) + 3.5(50) = 420 + 175 = 595$- At $B(20, 140): Z = 3(20) + 3.5(140) = 60 + 490 = 550$- At $C(40, 100): Z = 3(40) + 3.5(100) = 120 + 350 = 470$The maximum nitrogen is $595 \, kg$ at $x = 140$ and $y = 50$.

Similar Questions

The maximum value of the objective function $z=4x+6y$ subject to the constraints $3x+2y \leq 12$,$x+y \geq 4$,$x, y \geq 0$ is

The maximum value of $Z=10 x+25 y$ subject to $0 \leq x \leq 3, 0 \leq y \leq 3, x+y \leq 5, x \geq 0, y \geq 0$ is

The cost function $Z$ is given by $Z = 4x + 6y$. It is to be minimized. The feasible region for this function $Z$ is the shaded region represented in the following figure. Then the minimum value of $Z$ is and occurs at the point:

The maximum value of $Z = 4x + 5y$, subject to the constraints $3x + y \leq 15$, $3x + 4y \leq 24$, $x \geq 0$, $y \geq 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

	Brand $P$ ($kg$ per bag)	Brand $Q$ ($kg$ per bag)
Nitrogen	$3$	$3.5$
Phosphoric acid	$1$	$2$
Potash	$3$	$1.5$
Chlorine	$1.5$	$2$