 Hello and welcome to the session. I am Deepika here. Let's discuss the question which says one kind of cake requires 200 gram of flour and 25 gram of fat and another kind of cake requires 100 gram of flour and 50 gram of fat Find the maximum number of cakes which can be made from 5 kilogram of flour and 1 kilogram of Fat assuming that there is no shortage of the other ingredients used in making the cakes So let's start the solution Now here we will first formulate the linear programming problem according to the given conditions and then solve it graphically Now according to the question there are two kinds of cakes one kind of cake requires 200 gram of flour and 25 gram of fat and The another kind of cake requires 100 gram of flour and 50 gram of fat Now we have to find the maximum number of cakes which can be made from 5 kilogram of flour and 1 kilogram of fat Let the number of one kind of cake made is equal to x and the number of another kind of cake made is equal to y Now it is given that requirement of flour in making one kind of cake and another kind of cake is 200 gram and 100 gram respectively and the quantity of flour available is 5 kilogram that is 5000 grams Therefore, we have the first constraint as 200 x plus 100 y less than equal to 5000 So this is a constraint related to requirement of flour or this can be written as 2x plus y less than equal to 50 Also, we are given the requirement of fat in making one kind of cake is 25 gram and requirement of fat making in another kind of cake is 50 gram and quantity of fat available is 1 kilogram or 1000 gram So again, we have 25x Plus 50 y less than equal to 1000 So this is a constraint related to requirement of fat This can be written as x plus 2y less than equal to 40 Also, the number of cakes made is greater than equal to 0 Thus we have x greater than equal to 0 and y greater than equal to 0 so the mathematical formulation of the problem is maximized Z is equal to x plus y Subject to the constraints plus y less than equal to 50 x plus 2y less than equal to 40 x greater than equal to 0 and y greater than equal to 0 So this is our objective function. Let us give this as number one Now we have to maximize z is equal to x plus y Subject to these constraints. Let us give number to these constraints as two three and four So now we will draw the graph and find the feasible region subject to these given constraints Now equation of the line Corresponding to the inequality 2x plus y less than equal to 50 is 2x plus y is equal to 50 So we will first draw the line representing the equation 2x plus y is equal to 50 Now clearly the points 0 50 and 25 0 lie on the line 2x plus y is equal to 50 Therefore the graph of this line can be drawn by plotting points 0 50 and 25 0 and then joining them So let us take a as a point 0 50 and Be as the point 250 So a b is the line representing the equation 2x plus y is equal to 50 Now this line a b divides the plane into two half planes So we will consider the half plane which will satisfy two So the closed half plane containing the origin is the graph of two Now again the line corresponding to the inequality x plus 2y less than equal to 40 is x plus 2y is equal to 40 So we will draw the line representing the equation x plus 2y is equal to 40 on the same graph now the points 0 20 and 40 0 Satisfies the equation x plus 2y is equal to 40 So we will plot these points on the same graph and then join them Now, let us take c as a point 0 20 and d as a point 40 0 So CD represent the equation x plus 2y is equal to 40 Now again CD divides the plane into two half planes We will consider the half plane which satisfy 3 that is which satisfy the inequality x plus 2y less than equal to 40 So the half plane containing the origin is the graph of 3 Again x greater than equal to 0 and y greater than equal to 0 Implies that the graph lies in the first quadrant only Now here we observe that the lines a b and c g intersect at a point Let us take this point as p so the coordinates of p are 20 10 Now here the shaded portion in this graph is the feasible region Satisfying all the given constraints Now clearly the feasible region is bounded So we will use the corner point method to find the maximum value of set now the coordinates of corner points c o bpr 0 20 0 0 25 0 and 20 10 respectively Now according to the corner point method maximum value of set will occur at any of these points Therefore we will evaluate z at these points Now z is equal to x plus 5. So at the point see whose coordinates are 0 20 z is equal to 0 plus 20 which is equal to 20 again at origin z is equal to 0 and at the point b whose coordinates are 250 z is equal to 25 and At the point p whose coordinates are 20 10 z is equal to 20 plus 10 which is equal to 30 hence maximum value of z is equal to 30 which occurs when x is equal to 20 and y is equal to 10 therefore Maximum number of cakes that can be made from 5 kilogram of flour and 1 kilogram of fat is equal to 30 number of cakes of one kind is equal to 20 and number of cakes of another kind is equal to 10 Hence answer for this question is maximum number of cakes is equal to 30 20 cakes of kind 1 and 10 cakes of another kind So this completes our session. I Hope the solution is clear to you My and have a nice day