An open box with a square base is to be made out of a given quan

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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

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.


Let x be the side of the square base of the open box and y be its height. Let V denote the volume of the box.
    therefore space space space space straight V space equals space straight x squared straight y                                               ...(1)
Also surface area  = c2
rightwards double arrow space space space space space straight x squared plus 4 xy space equals space straight c squared space space space space space space rightwards double arrow space straight y space equals space fraction numerator straight c squared minus straight x squared over denominator 4 straight x end fraction
therefore space space space space from space left parenthesis 1 right parenthesis comma space space space space space straight V space equals straight x squared open parentheses fraction numerator straight c squared minus straight x squared over denominator 4 space straight x end fraction close parentheses
therefore space space space space space space straight V space equals space 1 fourth straight x left parenthesis straight c squared minus straight x squared right parenthesis space equals space 1 fourth left parenthesis straight c squared straight x minus straight x cubed right parenthesis
space space space space space space space dV over dx equals 1 fourth left parenthesis straight c squared minus 3 straight x squared right parenthesis
space space space space space space space space space space fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space 1 fourth left parenthesis negative 6 space straight x right parenthesis space equals space minus fraction numerator 3 straight x over denominator 2 end fraction
For V to be maximum or minimum,
                            dV over dx space equals space 0
therefore space space space space space 1 fourth left parenthesis straight c squared minus 3 straight x squared right parenthesis space equals space 0
rightwards double arrow space space space 3 straight x squared space equals straight c squared space space space space space space space space space space space space space space space space rightwards double arrow space space space straight x squared space equals straight c squared over 3
therefore space space space space straight x space equals space fraction numerator straight c over denominator square root of 3 end fraction space as space straight x space cannot space be space negative
For space straight x space equals space fraction numerator straight c over denominator square root of 3 end fraction comma space space space fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space minus 3 over 2. space fraction numerator straight c over denominator square root of 3 end fraction space equals space minus fraction numerator square root of 3 over denominator 2 end fraction space straight c space less than 0
therefore space space space straight V space is space maximum space when space straight x space equals space fraction numerator straight c over denominator square root of 3 end fraction
Maximum space value space of space straight V space equals space 1 fourth open parentheses straight c squared fraction numerator straight c over denominator square root of 3 end fraction minus fraction numerator straight c cubed over denominator 3 square root of 3 end fraction close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 fourth open parentheses fraction numerator straight c cubed over denominator square root of 3 end fraction minus fraction numerator straight c cubed over denominator 3 square root of 3 end fraction close parentheses space equals space 1 fourth fraction numerator 3 straight c cubed minus straight c cubed over denominator 3 square root of 3 end fraction space equals 1 fourth cross times fraction numerator 2 straight c cubed over denominator 3 square root of 3 end fraction space equals fraction numerator straight c cubed over denominator 6 square root of 3 end fraction cubic space units.

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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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