 I am welcome to the session, let us discuss the following question, question says find two positive numbers whose sum is 16 and some of whose cubes is minimum. First of all let us understand that if function f is defined on interval i and c is any point belonging to interval i such that f double dash c exist then if f dash c is equal to 0 and f double dash c is greater than 0 then c is a point of local minima and if f dash c is equal to 0 and f double dash c is less than 0 then c is a point of local maxima. This is the key idea to solve the given question. Now let us start the solution, let us assume that one of the positive number is equal to x, other positive number is equal to 16 minus x, we have given sum of two numbers is 16. So if one of the number is x then other number is equal to 16 minus x, now let p represents sum of their cubes that is x cube plus 16 minus x whole cube. Now differentiating both sides of this expression with respect to x we get dp upon dx is equal to 3x square plus 3 multiplied by 16 minus x whole square multiplied by minus 1, this implies dp upon dx is equal to 3x square minus 3 multiplied by 16 minus x whole square. Now we will find all the points at which dp upon dx is equal to 0, now dp upon dx is equal to this expression so we can write 3x square minus 3 multiplied by 16 minus x whole square equal to 0, now this implies 3x square is equal to 3 multiplied by 16 minus x whole square, we have transpose this term to right hand side, now dividing both sides by 3 we get x square is equal to 16 minus x whole square, now rearranging we can write it as 16 minus x whole square is equal to x square, taking square root on both the sides we get 16 minus x is equal to x on both the sides we have neglected the negative terms, we know both the numbers are positive numbers so here we have we have taken value of x is positive and we know sum of the two numbers is 16, so we will subtract one of the number from 16, we cannot subtract 16 from one of the number, so that is why we have neglected negative values on both the sides since both the numbers are positive, it is given in the question itself, now transposing the light terms we get 16 is equal to 2x, now dividing both sides by 2x is equal to 8, we get 8 is equal to x or we can write x equal to 8, now we get at x equal to 8 dp upon dx is equal to 0, now we know dp upon dx is equal to 3x square minus 3 multiplied by 16 minus x whole square, now differentiating both sides with respect to x again we get dp upon dx square equal to 6x minus 6 multiplied by 16 minus x multiplied by minus 1, now on simplifying we get 6x plus 6 multiplied by 16 minus x, now this is equal to 6x plus 96 minus 6x, now 6x and 6x will cancel each other and we get d square p upon dx square equal to 96, now we can write at x equal to 8, dp upon dx is equal to 0 and d square p upon dx square is equal to 96 which is greater than 0, so by key idea we get x is equal to 8 is a point of local minimum of p or we can write p equal to x cube plus 16 minus x whole cube is minimum at x equal to 8, now we get required two numbers are 8 and 16 minus 8, we know 16 minus 8 is equal to 8 only, so two numbers are 8 and 8, now this completes the session hope you understood the session take care and keep learning.