Let the diet contain x and y packets of foods P and Q respectively. Hence,
x ≥ 0 and y ≥ 0
The mathematical formulation of the given problem is given below
Maximize z = 6x + 3y
Subject to the constraints,
![\[4x\text{ }+\text{ }y\text{ }\ge \text{ }80\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-af25ff62a94423b90153ea6f62b97c68_l3.png "Rendered by QuickLaTeX.com")
![\[x\text{ }+\text{ }5y\text{ }\ge \text{ }115\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a6f581aaf8ffd9de91dadc9bc862b9e9_l3.png "Rendered by QuickLaTeX.com")
![\[3x\text{ }+\text{ }2y\text{ }\le \text{ }150\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c719dadffe7f3bdaf1738edd6aa03f47_l3.png "Rendered by QuickLaTeX.com")
x, y ≥ 0 …………… (v)
The feasible region determined by the system of constraints is given below
A (15, 20), B (40, 15) and C (2, 72) are the corner points of the feasible region
The values of z at these corner points are given below
| Corner point | z = 6x + 3y | |
| A (15, 20) | 150 | |
| B (40, 15) | 285 | Maximum |
| C (2, 72) | 228 |
So, the maximum value of z is 285 at (40, 15)
Hence, to maximize the amount of vitamin A in the diet, 40 packets of food P and 15 packets of food Q should be used.
The maximum amount of vitamin A in the diet is 285 units.