A square tank of capacity 250 cubic metres has to be dug out. Th

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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.

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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?


Let x side of square base.
 Volume of tank  = 250 cubic metres.
 therefore space space space space space space 250 space equals space straight x squared straight h space space space space space space space space space space space space space space space space space space rightwards double arrow space space space space straight h space equals space 250 over straight x squared                              ...(1)
Cost of land = Rs. space left parenthesis straight x squared space cross times space 50 right parenthesis space equals space Rs. space left parenthesis 50 space straight x squared right parenthesis
Cost of digging  = Rs. space left parenthesis 400 space straight h squared right parenthesis space equals space 400 space cross times space open parentheses 250 over straight x close parentheses squared comma space space space space space space space space space space space space space space space space space space space space space open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets
Let C be total cost.
  therefore space space space space space space space space straight C space equals space 50 straight x squared plus 400 space cross times open parentheses 250 over straight x squared close parentheses squared space equals space 50 straight x squared plus 25000000 over straight x to the power of 4
space space space space space space space space dC over dx space equals space 100 straight x minus 100000000 over straight x to the power of 5
Now space space space dC over dx space equals space 0 space space given space us
space space space space space 100 straight x space minus 100000000 over straight x to the power of 5 space equals 0 space space space space space space rightwards double arrow space space space straight x to the power of 6 space equals space 1000000
rightwards double arrow space space space space space space space space straight x space equals space 10
space space space space space space fraction numerator straight d squared straight C over denominator dx squared end fraction space equals space 100 plus 500000000 over straight x to the power of 6
space space space space space space space space
When         straight x space equals 10 comma space space fraction numerator straight d squared straight C over denominator dx squared end fraction space equals space 100 plus fraction numerator 500000000 over denominator left parenthesis 10 right parenthesis to the power of 6 end fraction space equals space 100 plus 500 space equals space 600 space greater than space 0
therefore space space space space space straight C space is space minimum space when space straight x space equals space 10
From space left parenthesis 1 right parenthesis comma space space space straight h space equals space fraction numerator 250 over denominator left parenthesis 10 right parenthesis squared end fraction space equals space 250 over 100 space equals space 2.5
therefore space space space square space tank space is space of space side space 10 space metres space and space height space 2.5 space metres.

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