 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says The corner points of the feasible region determined by the following system of linear inequalities 2x plus y less than equal to 10 x plus 3 y less than equal to 15 x and y greater than equal to 0 are 0 0 5 0 3 4 and 0 5 Let z is equal to p x plus q y where p and q greater than 0 Condition on p and q so that the maximum offset occurs at both 3 4 and 0 5 is p is equal to q p is equal to 2 q p is equal to 3 q q is equal to 3 p So let's start the solution Now we are given the corner points of the feasible region Determined by the linear inequality is 2x plus y less than equal to 10 x plus 3 y less than equal to 15 x greater than equal to 0 and y greater than equal to 0 are 0 0 5 0 3 4 and 0 5 Again the objective function given to us is z is equal to p x plus q y Therefore we will evaluate z at each corner point Now we have z is equal to p x plus q y Now at the point 0 0 z is equal to p into 0 plus q into 0 which is equal to 0 Now at the point 0 0 z is equal to p into 5 plus q into 0 which is equal to 5 p Again at the point 3 4 z is equal to p into 3 plus q into 4 and this is equal to 3 p plus 4 q Now at the point 0 5 z is equal to p into 0 plus q into 5 and this is equal to 5 q Now we have to condition on p and q so that the maximum of z occurs at both the points that is at the point 3 4 and at 0 5 So if maximum of z occurs both 3 4 and 0 5 then the value of z at the point 3 4 should be equal to value of z at the point 0 5 That is then 3 p plus 4 q is equal to 5 q or 3 p is equal to 5 q minus 4 q or 3 p is equal to q So the condition on p and q so that the maximum of z occurs at both the points that is at the point 3 4 and at 0 5 is q is equal to 3 p and this is our option d hence the answer for this question is d So this completes our session I hope the solution is clear to you Bye and have a nice day