 Hi and welcome to the session. Let us discuss the following question. Question says, find two positive numbers x and y such that their sum is 35 and the product of x square multiplied by y raised to the power 5 is a maximum. First of all let us understand that if we are given a function f defined on interval i, c belongs to interval i such that f double dash c exists. Then x is equal to c is a point of local maxima if f dash c is equal to 0 and f double dash c is less than 0. The value of fc is local maximum value of function f. This is the key idea to solve the given question. Let us now start the solution. We know sum of two positive numbers x and y is 35 so we can write x plus y is equal to 35. Now let us name x square multiplied by y raised to the power 5 as p. So we can write let p is equal to x square multiplied by y raised to the power 5. Now let us name this expression as 1 and this expression as 2. Now from 1 we get y is equal to 35 minus x. Now we will substitute this value of y in expression 2. Let us name this expression as 3. So we can write substituting value of y from expression 3 in expression 1. We get p is equal to x square multiplied by 35 minus x whole raised to the power 5. Now differentiating both sides with respect to x we get dp upon dx is equal to x square multiplied by 5. 35 minus x whole raised to the power 4 multiplied by minus 1 plus 35 minus x whole raised to the power 5 multiplied by 2x. Simplifying we get dp upon dx is equal to minus 5x square multiplied by 35 minus x whole raised to the power 4 plus 2x multiplied by 35 minus x whole raised to the power 5. Now we will find all the points at which dp upon dx is equal to 0. So we will put dp upon dx is equal to 0. Now this implies minus 5x square multiplied by 35 minus x whole raised to the power 4 plus 2x multiplied by 35 minus x whole raised to the power 5 is equal to 0. Now taking 35 minus x whole raised to the power 4 common on left hand side we get minus 5x square plus 2x multiplied by 35 minus x multiplied by 35 minus x whole raised to the power 4 is equal to 0. Now this implies 35 minus x whole raised to the power 4 multiplied by minus 5x square plus 70x minus 2x square is equal to 0. This implies 35 minus x whole raised to the power 4 multiplied by minus 7x square plus 70x is equal to 0. We know minus 5x square minus 2x square is equal to minus 7x square. Now we get 35 minus x whole raised to the power 4 is equal to 0 or minus 7x square plus 70x is equal to 0. Now here if we take 4 truth on both the sides we get 35 minus x is equal to 0. We get if we take minus 7x common on this side we get minus 7x multiplied by x minus 10 is equal to 0. Now here we get x is equal to 35. Here we get minus 7x is equal to 0 or x minus 10 is equal to 0. Now this implies x is equal to 35 or now if we divide both sides by minus 7 we get x is equal to 0. Here if we add 10 on both the sides we get x is equal to 10. So we get x is equal to 35 or x is equal to 0 or x is equal to 10. Now we know p is equal to x square multiplied by 35 minus x whole raised to the power 5. This we have already shown above. Clearly we can see at x is equal to 35 and at x is equal to 0 value of p is equal to 0. So we will neglect these two values of x. So we can write neglecting x is equal to 35 and x is equal to 0 we get x is equal to 10. Now to find if p is maximum and x is equal to 10 we will find second derivative of p with respect to x. We know dp upon dx is equal to this expression this we have already shown above. Now to find second derivative of p we will differentiate both the sides with respect to x again. So we get d square p upon dx square is equal to minus 5x square multiplied by 4 multiplied by 35 minus x whole cube multiplied by minus 1 plus 35 minus x whole raised to the power 4 multiplied by minus 10x. We will apply the product rule to find derivative of this term. Now again we will apply product rule to find derivative of this term. Now we can write 2x multiplied by 5 multiplied by 35 minus x whole raised to the power 4 multiplied by minus 1 plus 35 minus x whole raised to the power 5 multiplied by 2. Now simply find we get d square p upon dx square is equal to 35 minus x whole cube multiplied by 20x square minus 10x multiplied by 35 minus x minus 10x multiplied by 35 minus x plus 2 multiplied by 35 minus x whole square. Now this can be further written as d square p upon dx square is equal to 35 minus x whole cube multiplied by 20x square minus 20x multiplied by 35 minus x. You know these two are like terms so minus 10x multiplied by 35 minus x minus 10x multiplied by 35 minus x is equal to minus 20x multiplied by 35 minus x plus 2 multiplied by 35 minus x whole square. Now we will find the value of d square p upon dx square at x is equal to 10 this is equal to 35 minus 10 whole cube multiplied by 20 multiplied by 10 square minus 20 multiplied by 10 multiplied by 35 minus 10 plus 2 multiplied by 35 minus 10 whole square now this is further equal to 25 cube multiplied by 2000 minus 5000 plus 1250. Now this is further equal to 25 cube multiplied by minus 1750 which is less than 0. Now clearly we can see at x is equal to 10 dp upon dx is equal to 0 and d square p upon dx square is less than 0. So this implies x is equal to 10 is a point of local maxima or we can say p is maximum at x is equal to 10. We know required two numbers are x and y and x is equal to 10 and we also know that x plus y is equal to 35 this is given in the question. Now putting x is equal to 10 here we get y is equal to 35 minus 10 which is equal to 25 so value of x and y where p is maximum are x is equal to 10 and y is equal to 25. Hence the required two positive numbers are 10 and 25 this completes the session hope you understood the session take care and have a nice day.