A square piece of 30 cm is to be made into a box without top by

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 Multiple Choice QuestionsLong Answer Type

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?
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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

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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?


Let x (0 < x < 15) be the length of each side of the square which is to be cut from corners of the square tin sheet of each side 30 cm. Let V be the volume of the open box formed by folding up the flaps.

 therefore space space space space straight V space equals space straight x space left parenthesis 30 minus 2 straight x right parenthesis thin space left parenthesis 30 minus 2 straight x right parenthesis space equals space straight x left parenthesis 30 minus 2 straight x right parenthesis squared space equals space 4 straight x left parenthesis 15 minus straight x right parenthesis squared
space space space space space space space space space space space space equals space 4 straight x left parenthesis straight x squared minus 30 straight x plus 225 right parenthesis space equals space 4 left parenthesis straight x cubed minus 30 straight x squared plus 225 straight x right parenthesis
space space space space dV over dx space equals space 4 left parenthesis 3 straight x squared minus 60 straight x plus 225 right parenthesis
space space space space dV over dx space equals space 0 space space space space rightwards double arrow space space space space 4 left parenthesis 3 straight x squared minus 60 straight x plus 225 right parenthesis space equals space 0 space space space space space space rightwards double arrow space space space space space straight x squared minus 20 straight x plus 75 space equals space 0
rightwards double arrow space space space space space space left parenthesis straight x minus 5 right parenthesis thin space left parenthesis straight x minus 15 right parenthesis space equals space 0 space space space space rightwards double arrow space space space space space straight x space equals space 5 comma space space 15 space space space space space space space rightwards double arrow space space space space space straight x space equals space 5 space space space as space space straight x space equals space 15 space not an element of space left parenthesis 0 comma space 15 right parenthesis
space space space space space space space space fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space 4 left parenthesis 6 straight x minus 60 right parenthesis
At space space space straight x space equals space 5 comma space space space space space fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space 4 space space left parenthesis 30 minus 60 right parenthesis space equals space minus 120 space less than 0
therefore space space space space space space space space space space space straight V space space is space maximum space at space straight x space equals space 5 space space space space space space space space space space space space space space space space space space space space space space space
But x = 5 is the only extreme point
therefore space space space space space space straight V space is space maximum space at space straight x space equals space 5
therefore space space space space side space of space the space square space space equals space 5 space cm.
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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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