Question

A shopkeeper wishes to purchase a mmiber of 5 L oil tins and 1 kg ghee tins. He has only Rs.5760 to invest and has a. space to store at most 20 items. A 5 L oil tin costs him Rs.360 and a 1 kg ghee tin cost him Rs.240. His expectation is that he can sell an oil tin at a profit of Rs.22 and a ghee tin at a profit of Rs.18. Assuming that he can sell all the items he can buy, how should he invest his money in order to maximize the profit? Formulate this as a linear programming problem and solve it graphically.

Answer 1

Answer:

He should purchase 8 tins of 5 L oil and 12 tins of 1 kg Ghee.

Step-by-step explanation:

Let assume that shopkeeper purchase x tins of 5 L oil and y tins of 1 kg ghee.

It is given that, A 5 L oil tin costs him Rs.360 and a 1 kg ghee tin cost him Rs.240 and has to invest Rs 5760.

So, we have

$\sf \: 360x + 240y \leqslant 5760 - - - (1)$

Further given that, has a. space to store at most 20 items.

$\sf \: x + y \leqslant 20 - - - (2)$

Also, Objective function is to maximize the profit.

$\sf \: Max\:Z=22x + 18y$

So, we have now

$\sf \: Max\:Z=22x + 18y$

subject to the constraints

$\sf \: 360x + 240y \leqslant 5760 - - - (1)$

$\sf \: x + y \leqslant 20 - - - (2)$

$\sf \: x,y \geqslant 0$

Consider

$\sf \: x + y \leqslant 20$

Let we first sketch the line x + y = 20

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\qquad\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 20 \\ \\ \sf 20 & \sf 0 \end{array}} \\ \end{gathered}$

Now, Consider

$\sf \: 360x + 240y \geqslant 5760$

Let we first sketch the line 360x + 240y = 5760

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\qquad\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 24 \\ \\ \sf 16 & \sf 0 \end{array}} \\ \end{gathered}$

[ See the attachment graph ]

Now, from graph we concluded that OABC is a feasible region.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Corner \: Point & \bf Value \: of \: Z = 22x + 18y\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf A(16,0) & \sf 352\\ \\ \sf B(8,12) & \sf 392\\ \\ \sf C(0, 20) & \sf 360\\ \\ \sf O(0,0) & \sf 0 \end{array}} \\ \end{gathered}$

$\sf\implies \boxed{ \bf{ \:Maximum \: value \: of \: Z = 392 \: at \: (8,12)}}$

So, he should purchase 8 tins of 5 L oil and 12 tins of 1 kg Ghee.