📖 generic · CBSE Class 12th English Medium · MATHEMATCS PART-1 · Page 23question

A Note The reader may note that in Example 35, we have used first derivative · Part 13

Chapter 6: APPLICATION OF DERIVATIVES · MATHEMATCS PART-1

is to be constructed by removing equal squares from each corner of a metre by metre rectangular sheet of aluminium and folding up the sides. Find the volume of the largest such box. Solution Let x metre be the length of a side of the removed squares. Then, the height of the box is x , length is – x and breadth is – x (Fig .

). If V( x ) is the volume of the box, then Fig . V( x ) = x ( – x ) ( – x ) = x – x + x Therefore V ( ) ( )( ) V ( ) ′ ′′  Now V ′ ( x ) = gives , x = . But x ≠ (Why?) Thus, we have x = .

Now V ′′ = − Therefore, x = is the point of maxima, i.e., if we remove a square of side metre from each corner of the sheet and make a box from the remaining sheet, then the volume of the box such obtained will be the largest and it is given by V  = m Example Manufacturer can sell x items at a price of rupees     each. The cost price of x items is Rs     . Find the number of items he should sell to earn maximum profit. Solution Let S( x ) be the selling price of x items and let C( x ) be the cost price of x items.

Then, we have S( x ) = and C( x ) = x + Thus, the profit function P( x ) is given by P( x ) = S( ) C( ) i.e. P( x ) = x − P ′ ( x ) = Now P ′

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