Question

A firm manufactures two types of product, $a$ and $b$ , and sells them at a profit of $₹ 5$ per unit of type $a$ and $₹ 3$ per unit of type $b$ . Each product is processed on two machines, $M_{1}$ and $M_{2}$ . One unit of type $a$ requires one minute of processing time on $M_{1}$ and two minutes of processing time on $M_{2}$ , whereas one unit of type $b$ requires one minute of processing time on $M_{1}$ and one minute on $M_{2}$ . Machines $M_{1}$ and $M_{2}$ are respectively available for at most $5$ hours and $6$ hours in a day. Find out how many units of each type of product the firm should produce a day in order to maximise the profit. Solve the problem graphically.

Accepted Answer

Given data is tabulated as follows

Machines Time for type $a$ in minutes Time for type $b$ in minutes Available time in minutes

$M_{1}$ $1$ $1$ $300$

$M_{2}$ $2$ $M_{2}$ $360$

Profit per unit in rupees $5$ $5$

Let $x$ be number of $a$ and $y$ be number of $b$ products.

$∴$ According to the question,

$x + y \leq 300, 2 x + y \leq 360, x \geq 0, y \geq 0$

Let $Z$ be profit function.

Maximise $Z = 5 x + 3 y$

Consider, $2 x + y = 360$

at $₹ 3$

at $y = 0 \Rightarrow 2 x = 360 \Rightarrow x = 180$

Consider, $x + y = 300$

at $x = 0 \Rightarrow y = 300$

at $y = 0 \Rightarrow x = 300$

On solving $x + y = 300, 2 x + y = 360$ , we get $x = 60, y = 240$

The feasible region determined $x + y \leq 300, 2 x + y \leq 360, x \geq 0, y \geq 0$ is given by

The corner points of feasible region are $A (0, 0), B (0, 300), C (60, 240), D (180, 0)$ .

The value of $Z$ at corner point is

Corner Point $Z = 5 x + 3 y$

$A (0, 0)$ $0$

$B (0, 300)$ $900$

$C (60, 240)$ $1020$ Maximum

$D (180, 0)$ $900$

The maximum value of $Z$ is $1020$ and occurs at point $(60, 240)$ .

The firm should produce $60$ $a$ products and $240$ $b$ products to earn maximum profit of $R s . 1020$ .

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