Question

one kind of cake requires 300gm of flour and 15 gm of fat, another kind of cake requires 150gm of flour and 30 gm of fat. If there is 7,5 kg of flour and 600 gm of fat, assuming that there is no shortage of other ingredients used in making the cakes. Based on above information answer the following: (a) Write an algebraic expression of objective function. (b) Write the constraint for cake of type I. (c) Write the constraint for cake of type II. (d) Write the corner points of feasible region for given situation. (e) What will be the maximum number of cakes?

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that number of cakes of first kind and second kind that can be made be x and y respectively.

The Mathematical Formulation of the problem is as follow

$\red{\bf :\longmapsto\:Maximise \: Z = x + y}$

subject to the constraints,

$\rm :\longmapsto\:300x + 150y \leqslant 7500 - - (1)$

$\rm :\longmapsto\:15x + 30y \leqslant 600 - - (2)$

$\rm :\longmapsto\:x \geqslant 0 - - - (3)$

$\rm :\longmapsto\:y \geqslant 0 - - - (4)$

Now,

↝ Let us plot the graph of the system of inequalities from 1 to 4.

Consider,

$\rm :\longmapsto\:300x + 150y = 7500 - - (1)$

On substituting x = 0, we get

$\rm :\longmapsto\:0 + 150y = 7500$

$\rm :\longmapsto\:y = 50$

On substituting y = 0, we get

$\rm :\longmapsto\:300x + 0 = 7500$

$\rm :\longmapsto\:x = 25$

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 50 \\ \\ \sf 25 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 50) & (25 , 0)

➢ See the attachment graph.

Consider,

$\rm :\longmapsto\:15x + 30y = 600 - - (2)$

On substituting x = 0, we get

$\rm :\longmapsto\:0 + 30y = 600$

$\rm :\longmapsto \: 30y = 600$

$\rm :\longmapsto \: y = 20$

On substituting y = 0, we get

$\rm :\longmapsto\:15x + 0 = 600$

$\rm :\longmapsto\:15x= 600$

$\rm :\longmapsto\:x= 40$

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 20 \\ \\ \sf 40 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 20) & (40 , 0)

➢ See the attachment graph.

From the graph we concluded that OABC is the feasible region which is closed and bounded.

The coordinates of corner point of feasible region and corresponding value of Z is given below :

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Corner \: Point & \bf Z = x + y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf O(0,0) & \sf 0 \\ \\ \sf A(25, 0) & \sf 25\\ \\ \sf B(20,10) & \sf 30\\ \\ \sf C(0,20) & \sf 20 \end{array}} \\ \end{gathered}$

Hence,

↝ Maximum number of cakes of first kind is 20 and second kind is 10.