NCERT Class 12-Mathematics: Chapter –6 Application of Derivatives Part 19

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Question 30:

Find the dimensions of the rectangle of perimeter which will sweep out a volume as large as possible, when revolved about one of its sides. Also find the maximum volume.

Answer:

Given: rectangle of perimeter

To find: the dimensions of the rectangle which will sweep out a volume as large as possible, when revolved about one of its sides. Also to find the maximum volume

Explanation: Let the length and the breadth of the rectangle be and .

Then it is given perimeter of the rectangle is ,

Now when the rectangle revolve about side y it will form a cylinder with y as the height and x as the radius, then if the volume of the cylinder is V, then we know

Substituting value from equation (i) in above equation we get

Now we will find first derivative of the above equation, we get

Taking out the constant terms and applying the sum rule of differentiation, we get

Now to find the critical point we will equate the first derivative to 0, i.e.,

Now we will find the second derivative of the volume equation, this can be done by again differentiating equation (ii), we get

Taking out the constant terms and applying the sum rule of differentiation, we get

Now substituting (from equation (iii)), we get

Hence at will have maximum value.

The maximum value of can be found by substituting in , i.e.,

Hence the dimensions of the rectangle which will sweep out a volume as large as possible, when revolved about one of its sides equal to .

Question 31:

If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?

Answer:

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