 Hi and how are you all today? My name is Priyanka and I shall be helping you with the following question. It says the total cost and demand function of an item are given by c is equal to x cube upon 3 minus 7x square plus 111x plus 50 and p is equal to 100 minus x respectively. Find the total revenue function and the profit function. Find the number of items when the profit will be maximum. Find the maximum profit also. So here we are given the cost function as x cube upon 3 minus 7x square plus 111x plus 50 and we are given the demand function p is equal to 100 minus x. So with the help of this demand function we can easily find out the revenue function. Revenue function is equal to when price is multiplied by the quantity which is sold. So it is 100 minus x into x that is 100x minus x square. So this is our total revenue function that is 100x minus x square. Further we need to find out the profit function. We can find out profit function by subtracting cost function from the revenue function. This is our revenue function right? So we have 100x minus x square minus cost function that is x cube upon 3 minus 7x square plus 111x plus 50. So it is equal to 100x minus x square minus x cube upon 3 plus 7x square minus 111x minus 50. Simplifying it further we have a positive 100 and a negative 111. So it is firstly in order minus x cube upon 3 plus 6x square minus 11x minus 50. So this is our profit function. Further we need to find out the number of items when the profit will be maximum. We know that profit is maximum when the first derivative of this profit function with respect to x is equal to 0 and the second derivative is less than 0. So let us find out the first derivative of this profit function equated to 0. Find out the value of x and then find out the second derivative. Substitute the value of x in that and find out whether it is less than 0 or not. So let us find out the first derivative of the profit function. We are given the profit function that is we have found out the profit function as minus x cube upon 3 plus 6x square minus 11x minus 50. So let us differentiate this profit function with respect to x. So we have minus 3x square upon 3 plus 12x minus 11. That is further equal to minus x square plus 12x minus 11. This is the first derivative. Now equate it to 0 and find out the value of x. We have minus x square plus 12x minus 11 equal to 0 now. That implies x square minus 12x plus 11 is equal to 0. This further implies x minus 1 into x minus 11 is equal to 0. This further implies x is equal to 1 or it is equal to 11. Now we will find out the second derivative also. We have d square e upon dx square equal to d by dx of the first derivative. So it is d by dx of minus x square plus 12x minus 11. This is the first derivative. This is further equal to minus 2x plus 12. Now when x is equal to 1 then d square p upon dx square will be minus 2 plus 12. That is a positive 10 which is greater than 0. So that means this is not the value of x. Now when x is equal to 11 then we have d square p upon dx square equal to minus 22 plus 12 which is minus 10 which is less than 0. So this implies that profit is maximum when x is equal to 11. Lastly we need to find out the maximum profit also. So what we will do is we will put the value of x that is 11 in profit function. We found out above our profit function as minus x cube upon 3 plus 6x square minus 11x minus 50 right. When x is equal to 11 it will be minus 11 cube upon 3 plus 611 square minus 11 into 11 minus 50 which is further equal to rupees 111.33 right. So this is our maximum profit that is rupees 111.33 pesos right. So this completes the whole session. Hope you understood. Remember to find out the maximum profit also as it is mentioned in the question right. So have a nice day.