A diet is to contain at least 80 units of vit | vedclass

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Question

(A) Let the diet contain $x$ units of food $F_{1}$ and $y$ units of food $F_{2}$.The given information can be compiled in a table as follows:FoodVitam

Vedclass · Accepted Answer

$104$

Vedclass · Answer

(A) Let the diet contain $x$ units of food $F_{1}$ and $y$ units of food $F_{2}$.The given information can be compiled in a table as follows:FoodVitamin $A$ (units)Minerals (units)Cost (Rs)$F_{1} (x)$$3$$4$$4$$F_{2} (y)$$6$$3$$6$Requirement$80$$100$-The mathematical formulation is:Minimize $Z = 4x + 6y$ subject to:$3x + 6y \geq 80$$4x + 3y \geq 100$$x, y \geq 0$The corner points of the feasible region are $A(\frac{80}{3}, 0)$,$B(24, \frac{4}{3})$,and $C(0, \frac{100}{3})$.Evaluating $Z$ at corner points:- At $A(\frac{80}{3}, 0)$,$Z = 4(\frac{80}{3}) + 6(0) = \frac{320}{3} \approx 106.67$- At $B(24, \frac{4}{3})$,$Z = 4(24) + 6(\frac{4}{3}) = 96 + 8 = 104$- At $C(0, \frac{100}{3})$,$Z = 4(0) + 6(\frac{100}{3}) = 200$Since the feasible region is unbounded,we check if $4x + 6y < 104$ has any common points with the feasible region. The line $4x + 6y = 104$ (or $2x + 3y = 52$) does not intersect the feasible region. Thus,the minimum cost is $Rs. 104$.

From/To	$A$	$B$	$C$
$P$	$160$	$100$	$150$
$Q$	$100$	$120$	$100$

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