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A diet is to contain at least 80 units of Vitamin A and 100 units of minerals. Two foods F1 and F2 are available costing Rs 5 per unit and Rs 6 per unit respectively. One unit of food F1 contains 4 units of vitamin A and 3 units of minerals whereas one unit of food F2 contains 3 units of vitamin A and 6 units of minerals. Formulate this as a linear programming problem. Find the minimum cost of a diet that consists of a mixture of these two foods and also meets minimum nutritional requirements.

 

 

 

 
 
 
 
 

Answers (1)

Let \mathrm{x} and \mathrm{y} in (units) of food F1 and F2 be mixed.

To minimize: \mathrm{z = 5x + 6y} in Rs.

Subject to constraints:    

\begin{matrix} x \geq 0 \ y \geq 0 & 4x + 3y \geq 80 \\ 3x +6y \geq 100 \end{matrix}

\begin{matrix} \underline{\text{Corner Points}} & \underline{\text{Value of z (in Rs.)}}\\A\left(0,\frac{80}{3} \right ) & 160 \\ B\left(12,\frac{32}{3} \right ) & 124 & \longleftarrow \text{ Min. Value} \\ C\left(100,0\right ) & \frac{500}{3}\end{matrix}

Since feasible region is unbounded so, 124 may or may not be minimum value of z.

To check, draw \mathrm{ 5x + 6y< 124}. As in the half-plane \mathrm{ 5x + 6y< 124}, there is not a point common with the feasible region.

 

Posted by

Ravindra Pindel

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