 Hi and welcome to the session. I'm Shashi and I'm going to help you with the following question. Question says, find two positive numbers x and y such that x plus y is equal to 60 and xy cube is maximum. First of all let us understand that if function f is defined on interval i, c is any point belonging to interval i such that f double dash c exist. Then c is a point of local maxima, f dash c is equal to 0 and f double dash c is less than 0 and c is a point of local minima is f dash c is equal to 0 and f double dash c is greater than 0. This is the key idea to solve the given question. Let us start the solution. We have given two positive numbers x and y such x plus y is equal to 60 and xy cube is maximum. Let p is equal to xy cube. Now let us name this expression as 1 and this expression as 2 from 1 we get. If we subtract x from both sides we get y is equal to 60 minus x. Now let us name this expression as 3. Now substituting the value of phi from 3 into we get p is equal to x multiplied by 60 minus x whole cube. Now differentiating both sides with respect to x we get dp upon dx is equal to here we will apply the product truth x multiplied by 3 multiplied by 60 minus x whole square multiplied by minus 1 plus 60 minus x whole cube multiplied by 1. Now simplifying we get dp upon dx equal to minus 3x multiplied by 60 minus x whole square plus 60 minus x whole cube. Now we will find all the points at which dp upon dx is equal to 0. Now we know dp upon dx is equal to minus 3x multiplied by 60 minus x whole square plus 60 minus x whole cube. So we can write minus 3x multiplied by 60 minus x whole square plus 60 minus x whole cube equal to 0. Now taking 60 minus x whole square common from both the terms we get 60 minus x whole square multiplied by minus 3x plus 60 minus x is equal to 0. This implies 60 minus x whole square multiplied by 60 minus 4x we know minus 3x minus x is equal to minus 4x. So we get 60 minus x whole square multiplied by 60 minus 4x equal to 0. This implies 60 minus x whole square is equal to 0 or 60 minus 4x is equal to 0. This implies here if we take square root on both the sides we get 60 minus x equal to 0 and here if we transpose the light terms we get 4x equal to 60. Now transposing the light terms we get x equal to 60 or now dividing both sides by 4 here we get x equal to 15. Now we get x equal to 60 and x equal to 15. At x equal to 60 y is equal to 0 we know x plus y is equal to 60 and we also know that x and y are two positive numbers. So we will reject this value x equal to 60. Now let us find out second derivative of p. Now we know dp upon dx is equal to minus 3x multiplied by 60 minus x whole square plus 60 minus x whole cube. Now differentiating both sides with respect to x we get d square p upon dx square is equal to here we will apply the product rule minus 3x multiplied by 2 multiplied by 60 minus x multiplied by minus 1 plus 60 minus x whole square multiplied by minus 3 plus now we will take the derivative of 60 minus x whole cube derivative of 60 minus x whole cube is 3 multiplied by 60 minus x whole square multiplied by minus 1. Now this is equal to 6x multiplied by 60 minus x minus 3 multiplied by 60 minus x whole square minus 3 multiplied by 60 minus x whole square d square p upon dx square equal to 6x multiplied by 60 minus x minus 6 multiplied by 60 minus x whole square. Let's find out the value of d square p upon dx square at x equal to 15 is equal to 6 multiplied by 15 multiplied by 60 minus 15 minus 6 multiplied by 60 minus 15 whole square. This is equal to 6 multiplied by 15 multiplied by 45 minus 6 multiplied by 45 square. This is equal to 6 multiplied by 45 multiplied by 15 minus 45. Here we have taken 6 multiplied by 45 common in both the terms. Now this is equal to 270 multiplied by minus 30. Now this is equal to minus 81 100. Now clearly we can see at x equal to 15 dp upon dx is equal to 0 and d square p upon dx square is equal to minus 81 100 which is less than 0. This implies x is equal to 15 is a point of local maxima and we can say p is maximum at x equal to 15. Now we get required two numbers are x equal to 15 and y equal to 60 minus 15 which is equal to 45. So this is our required answer. This completes the session. Hope you understood the session. Keep smiling and take care.