A dietician has to develop a special diet usi | vedclass

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

Question

(A) Let the diet contain $x$ and $y$ packets of foods $P$ and $Q$ respectively. The objective is to maximize $Z = 6x + 3y$. The constraints are: $12x

Vedclass · Accepted Answer

$40$ packets of $P$ and $15$ packets of $Q$; Maximum vitamin $A = 285$ units

Vedclass · Answer

(A) Let the diet contain $x$ and $y$ packets of foods $P$ and $Q$ respectively. The objective is to maximize $Z = 6x + 3y$. The constraints are: $12x + 3y \geq 240 \implies 4x + y \geq 80$ $4x + 20y \geq 460 \implies x + 5y \geq 115$ $6x + 4y \leq 300 \implies 3x + 2y \leq 150$ $x, y \geq 0$. Solving the intersection points of the lines: $4x + y = 80$ and $x + 5y = 115$ gives $A(15, 20)$. $4x + y = 80$ and $3x + 2y = 150$ gives $B(2, 72)$. $x + 5y = 115$ and $3x + 2y = 150$ gives $C(40, 15)$. Evaluating $Z = 6x + 3y$ at corner points: Corner Point$Z = 6x + 3y$$A(15, 20)$$6(15) + 3(20) = 150$$B(2, 72)$$6(2) + 3(72) = 228$$C(40, 15)$$6(40) + 3(15) = 285$ The maximum value is $285$ at $(40, 15)$. Thus,$40$ packets of $P$ and $15$ packets of $Q$ are required.

Similar Questions

The minimum value of $Z = 3x + y$, subject to the constraints $2x + 3y \leq 6$, $x + y \geq 1$, $x \geq 0$, and $y \geq 0$ is:

The maximum value of $z$ in the following equation $z=6xy+y^2$,subject to the constraints $3x+4y \leq 100$,$4x+3y \leq 75$,$x \geq 0$,and $y \geq 0$ is:

The optimal solution of the $L$.$P$.$P$. Maximize $Z = 8x + 3y$ subject to the constraints $x + y \leq 3, 4x + y \leq 6, x \geq 0, y \geq 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module