A wire of length 30 cm is to be cut into two pieces. One of the

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 Multiple Choice QuestionsLong Answer Type

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311. A wire of length 30 cm is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What could be the length of the two pieces, so that the combined area of the square and the circle is minimum?


Total length of wire = 30 metre. Let x metres be made into a circle and (30 – x) metres into a square.
therefore space space space radius space of space circle space space equals space fraction numerator straight x over denominator 2 space straight pi end fraction space metres space and space side space of space square space equals space fraction numerator 30 minus straight x over denominator 4 end fraction metres
Let S be the sum of areas of two figures. 
   therefore space space space space straight S space equals space straight pi open parentheses fraction numerator straight x over denominator 2 space straight pi end fraction close parentheses squared space plus space open parentheses fraction numerator 30 minus straight x over denominator 4 end fraction close parentheses squared space equals space fraction numerator straight x squared over denominator 4 straight pi end fraction plus fraction numerator left parenthesis 30 minus straight x right parenthesis squared over denominator 16 end fraction

              dS over dx space equals fraction numerator 2 straight x over denominator 4 straight pi end fraction plus fraction numerator 2 left parenthesis 30 minus straight x right parenthesis thin space left parenthesis negative 1 right parenthesis over denominator 16 end fraction space equals space fraction numerator straight x over denominator 2 straight pi end fraction minus fraction numerator 30 minus straight x over denominator 8 end fraction
dS over dx space equals 0 space space space gives space us space fraction numerator straight x over denominator 2 straight pi end fraction minus fraction numerator 30 minus straight x over denominator 8 end fraction equals 0

therefore space space space space 4 straight x minus 30 space straight pi space plus space straight x space straight pi space equals space 0 space space space space space rightwards double arrow space space space left parenthesis straight pi plus 4 right parenthesis space straight x space equals space 30 space straight pi
therefore space space space space space space space space space space space space space straight x space equals space fraction numerator 30 space straight pi over denominator straight pi space plus 4 end fraction
space space space space space space space space fraction numerator straight d squared straight S over denominator dx squared end fraction space equals space fraction numerator 1 over denominator 2 straight pi end fraction plus 1 over 8
When space straight x space equals space fraction numerator 30 space straight pi over denominator straight pi plus 4 end fraction comma space space space space space fraction numerator straight d squared straight S over denominator dx squared end fraction space equals space fraction numerator 1 over denominator 2 straight pi end fraction plus 1 over 8 greater than 0
therefore space space space space space space straight S space is space minimum space when space straight x space equals fraction numerator 30 space straight pi over denominator straight pi space plus 4 end fraction
therefore space space space space space space the space wire space is space to space be space cut space at space straight a space distance space of space fraction numerator 30 space straight pi over denominator straight pi plus 4 end fraction space meteres space from space one space end. space

        
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312.

The sum of the perimeters of a circle and square is k where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle. 

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 Multiple Choice QuestionsShort Answer Type

313. A square piece of 30 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. What should be the side of the square to be cut off so that volume of the box is maximum?
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314. A square piece of a tin of side 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 a box. What should be the side of the square to be cut off, so that the volume of the box is the maximum possible?
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315. A square piece of tin of side 24 cm. is to be made into a box without top by cutting a square from each corner arid folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum ? Also, find this maximum volume.
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316.

An open box with a square base is to be made out of a given quantity of sheet of area c2. Show that the maximum volume of the box is fraction numerator straight c cubed over denominator 6 square root of 3 end fraction.

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 Multiple Choice QuestionsLong Answer Type

317.

An open topped box is to be constructed by removing equal squares from each corner of a 3 metre by 8 metre rectangular sheet of aluminum and folding up the sides. Find the volume of the largest such box.

 
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 Multiple Choice QuestionsShort Answer Type

318.

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off squares from each corner and folding up the flaps. What should be side of the square to be cut off so that the volume of he box is maximum?

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 Multiple Choice QuestionsLong Answer Type

319. An open box with a square base is to be made out of a given iron sheet of area 27 square m. Show that the maximum volume of the box is 13.5 cubic m.
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320.

A square tank of capacity 250 cubic metres has to be dug out. The cost of the land is Rs.50 per square metre. The cost of digging increases with the depth and for the whole tank is Rs 400 h2, where h metres is the depth of the tank. What should be the dimensions of the tank so that the cost be minimum?

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