 Hi and welcome to the session. I am Shrashen and I am going to help you with the following question. Question says of all the closed cylindrical cans of a given volume of 100 cubic centimeters, find the dimensions of the can which has the minimum surface area. Now let us start with the solution. Let us assume that r is the radius and h is the height of the cylindrical can. We know volume of the cylindrical can is 100 cubic centimeters. Now we can write volume v of the cylindrical can is equal to 100. We also know that volume of a cylinder is equal to pi r square h where r is the radius of the cylinder and h is the height. So this implies pi r square h is equal to 100. Now this further implies h is equal to 100 upon pi r square dividing both sides by pi r square we get h is equal to 100 upon pi r square. Let surface area of the given cylindrical can be s. Now we know surface area of cylinder is given by 2 pi r h plus 2 pi r square. Now let us name this expression as 1 and this expression as 2. Now substituting the value of h from expression 1 in expression 2 we get s is equal to 2 pi r multiplied by 100 upon pi r square plus 2 pi r square. Now pi and pi will cancel each other and r will cancel 1 r. So we get 200 upon r plus 2 pi r square is equal to s. Now differentiating both sides with respect to r we get ds upon dr is equal to 200 upon r square multiplied by minus 1 plus 4 pi r. We can simply write it as ds upon dr is equal to 4 pi r minus 200 upon r square. We know minus 1 multiplied by 200 upon r square is minus 200 upon r square. Now we will find all the values of r where ds upon dr is equal to 0. So we will put ds upon dr is equal to 0. Now this implies 4 pi r minus 200 upon r square is equal to 0. Now this implies 4 pi r cube minus 200 upon r square is equal to 0. We have subtracted these 2 terms by taking their ncm. Now multiplying both sides by r square we get 4 pi r cube minus 200 is equal to 0. Now adding 200 on both sides we get 4 pi r cube is equal to 200. Now dividing both sides by 4 pi we get r cube is equal to 200. Now this implies r is equal to 200 upon 4 pi r is equal to 200 upon 4 pi r cube is equal to 200 upon 4 pi r square. Now this implies r is equal to cube root of we know 4 and 200 will cancel each other. We get 50 upon pi r cube is equal to 50 upon pi r square. Now we will find second derivative of s with respect to r. We know ds upon dr is equal to 4 pi r minus 200 upon r square. This we have already shown above. Now to find second derivative of s with respect to r we will differentiate both sides with respect to r again. We get d square s upon dr square is equal to 4 pi plus 400 upon r cube. Now let us find out the value of d square s upon dr square at r is equal to cube root of 50 upon pi. Now this is equal to 4 pi plus 400 upon cube root of 50 upon pi whole cube. Now this cube and cube root will cancel each other and we get 4 pi plus 400 upon 50 upon pi. Simplifying we get 4 pi plus 400 multiplied by pi upon 50. Now further simplifying we get 4 pi plus 8 pi we know 50 multiplied by 8 is equal to 400. Now this is equal to 12 pi 4 pi plus 8 pi is equal to 12 pi. Now this is greater than 0. Now from above discussion we know at r is equal to cube root of 50 upon pi ds upon dr is equal to 0 and d square s upon dr square is greater than 0. So this implies surface area is minimum at r is equal to cube root of 50 upon pi. Now to find the dimensions of height that is h we will use the expression 1. Now h is equal to 100 upon pi r square implies h is equal to 100 upon pi multiplied by cube root of 50 upon pi whole square. We know surface area is minimum at cube root of 50 upon pi. So here we have substituted the value of r. Now this implies h is equal to 100 upon pi multiplied by 50 upon pi raised to the power 2 upon 3. We know cube root of 50 upon pi can be written as 50 upon pi raised to the power 1 upon 3. That is what we have done here. Now this implies h is equal to 100 upon pi multiplied by 50 raised to the power 2 upon 3 upon pi raised to the power 2 upon 3. We know the rule x upon y raised to the power m can be written as x raised to the power m upon y raised to the power m. That is what we have done here. Now this implies h is equal to 100 divided by pi upon pi raised to the power 2 upon 3 multiplied by 50 raised to the power 2 upon 3. Now this implies h is equal to 100 upon pi raised to the power 1 minus 2 upon 3 multiplied by 50 raised to the power 2 upon 3. Here we can see base is same. So we will subtract the powers. So we get h is equal to 100 upon pi raised to the power 1 upon 3. We know 1 minus 2 upon 3 is equal to 1 upon 3 multiplied by 50 raised to the power 2 upon 3. Now we can write 100 as 2 multiplied by 50. So we get h is equal to 2 multiplied by 50 upon pi raised to the power 1 upon 3 multiplied by 50 raised to the power 2 upon 3. Now this can be further written as 2 divided by pi raised to the power 1 upon 3 multiplied by 50 raised to the power 1 minus 2 upon 3 where bases are same. So powers will get subtracted. So we get h is equal to 2 upon 5 raised to the power 1 upon 3 multiplied by 50 raised to the power 1 upon 3 you know 1 minus 2 upon 3 is equal to 1 upon 3. Now we can write it as h is equal to 2 multiplied by 50 upon 5 raised to the power 1 upon 3. This is further equal to 2 multiplied by cube root of 50 upon 5. So for the surface area to be minimum these are the required dimensions for radius and height. This completes the session. Hope you understood the session. Take care and have a nice day.