MHT CET · Maths · Linear Programming
A diet of a sick person must contain atleast 4000 unit of vitamins, 50 unit of proteins and 1400 calories. Two foods \(\mathrm{A}\) and \(\mathrm{B}\) are available at cost of ₹ \(4\) and ₹ \(3\) per unit respectively. If one unit of A contains 200 unit of vitamins, 1 unit of protein and 40 calories, while one unit of food B contains 100 unit of vitamins, 2 unit of protein and 40 calories. Formulate the problem, so that the diet be cheapest.
- A \(200 x+100 y \geq 4000, x+2 y \geq 50\)
\(40 x+40 y \geq 1400, x \geq 0\) and \(y \geq 0\)
\(O . F z=4 x+3 y\) - B \(400 x+200 y \geq 100, x+2 y \geq 50\)
\(40 x+40 y \geq 1400, x \geq 0\) and \(y \geq 0\)
O. \(F z=4 x+3 y\) - C \(100 x+200 y \geq 4000, x+2 y \geq 50\),
\(40 x+40 y \geq 1400, x \geq 0\) and \(y \geq 0\)
O. \(F z=4 x+3 y\) - D None of the above
Answer & Solution
Correct Answer
(A) \(200 x+100 y \geq 4000, x+2 y \geq 50\)
\(40 x+40 y \geq 1400, x \geq 0\) and \(y \geq 0\)
\(O . F z=4 x+3 y\)
Step-by-step Solution
Detailed explanation

Let \(z\) be the profit function and \(x\) and \(y\) denote the productivity of food \(A\) and \(B\) respectively. Then
\(200 x+100 y \geq 4000 \)
\( x+2 y \geq 50 \)
\( 40 x+40 \geq 1400 \)
\( \text {O.F } z=4 x+3 y, \quad x \geq 0 \text { and } y \geq 0\)
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