 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question says, a square piece of tin of site 18 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible? First of all let us understand that if we are given a function f defined on interval i and c belongs to interval i such that f double dash c exists then c is a point of local maxima if f dash c is equal to 0 and f double dash c is less than 0. This is the key idea to solve the given question. Now let us start the solution. Now let each side of square cut off from each corner be x cm. This is a square piece of tin. Each of its side is equal to 18 cm. We have to make a box without top by cutting off square from each corner. Let the side of the square be x cm. Now clearly we can see length and breadth of the box is same that is 18 minus 2 x cm. We will subtract this x and this x from 18. We know this complete side is equal to 18 cm. So we get side of the box as 18 minus 2 x cm. So we can write length of the box is equal to 18 minus 2 x cm and breadth of the box is also equal to 18 minus 2 x cm. Now clearly we can see height of the box is equal to x cm. This square has each side equal to x cm. So we get height as x cm. On folding of the flaps box will appear like this. Length and breadth are same and height is equal to x cm. Also box is open at the top. Now we can find the volume of the box by using the formula length into breadth into height. We know volume of a keyboard is equal to length into breadth into height. So we can write volume is equal to 18 minus 2 x cm multiplied by 18 minus 2 x cm multiplied by x cm or we can write it as 18 minus 2 x cm whole square multiplied by x cm. 18 minus 2 x cm multiplied by 18 minus 2 x cm can be written as 18 minus 2 x cm whole square. Now differentiating both sides with respect to x we get dv upon dx is equal to 2 multiplied by x 18 minus 2 x multiplied by minus 2 plus 18 minus 2 x whole square multiplied by 1. Here we have applied the product rule. Now this is further equal to minus 4 x multiplied by 18 minus 2 x plus 18 minus 2 x whole square. Now we will find all the points at which dv upon dx is equal to 0. This implies minus 4 x multiplied by 18 minus 2 x plus 18 minus 2 x whole square is equal to 0. Now this implies 18 minus 2 x multiplied by minus 4 x plus 18 minus 2 x is equal to 0. Here we have taken 18 minus 2 x common from both the terms. Now this is further equal 18 minus 2 x multiplied by 18 minus 6 x equal to 0. Now this implies 18 minus 2 x is equal to 0 or 18 minus 6 x is equal to 0. This implies minus 2 x is equal to minus 18. Here we have subtracted 18 from both the sides or again subtracting 18 from both the sides we get minus 6 x is equal to minus 18. Now this implies dividing both sides by minus 2 we get x is equal to 9 or now here we can divide both sides by minus 6. Now dividing both sides by minus 6 we get x is equal to 3. So we get x is equal to 9 or x is equal to 3. Now at x equal to 9 length and breadth both are equal to 0. So we will reject this value. Now neglecting x equal to 9 we get x is equal to 3. Now to find if x is equal to 3 is a point of local maxima we will find second derivative of v. Now we will differentiate dv upon dx with respect to x again. Now differentiating both sides with respect to x we get d square v upon dx square is equal to where we will apply product rule we get minus 4 x multiplied by minus 2 plus 18 minus 2 x multiplied by minus 4 plus 2 multiplied by 18 minus 2 x multiplied by minus 2. Here we will apply the chain rule to find the derivative of this term. Now this implies d square v upon dx square is equal to 8x minus 72 plus 8x minus 4 multiplied by 18 minus 2 x. This implies d square v upon dx square is equal to 8x plus 8x is equal to 16x minus 72 minus 72 plus 8x. We know minus 4 multiplied by 18 is equal to minus 72 and minus 4 multiplied by minus 2 x is equal to plus 8x. Now this implies d square v upon dx square is equal to 16x plus 8x is equal to 24x minus 144 minus 72 minus 72 is equal to minus 144. Now let us find out value of d square v upon dx square at x equal to 3. You can write at x equal to 3 d square v upon dx square is equal to 24 multiplied by 3 minus 144. This is equal to 72 minus 144. We know 24 multiplied by 3 is equal to 72. Now 72 minus 144 is equal to minus 72. So we get d square v upon dx square equal to minus 72 which is less than 0. Now we can write at x equal to 3 d v upon dx is equal to 0 and d square v upon dx square is equal to minus 72 which is less than 0. This implies x is equal to 3 is a point of local maxima. Here we have used the key idea. Now we can say volume is maximum at x is equal to 3. Now we get the required value of x is 3 centimeter. So we can write site of the square to be cut off from each corner is equal to 3 centimeter. So this is our required answer. This completes the session. Hope you enjoyed the session. Take care and have a nice day.