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Long Answer Type

331. A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m^3. If building of tank costs Rs. 70 per sq. metres for the base and Rs. 45 per square metre for sides. What is the cost of least expensive tank ?

Let x metres be length and y metres be breadth of base.
Now length of tank = 2 metres
Let V be volume of tank

therefore space space space space space straight V space equals 2 xy

But

straight V space equals space 8 space straight m cubed

(given)

therefore space space 2 xy space equals 8 space space space space rightwards double arrow space space space xy space equals 4 space space space space rightwards double arrow space space space straight y space equals space 4 over straight x

...(1)
Area of base = xy = 4m²

open square brackets because space of space left parenthesis 1 right parenthesis close square brackets

Area of four sides =

left parenthesis 2 straight x plus 2 straight y plus 2 straight x plus 2 straight y right parenthesis straight m squared space equals space left parenthesis 4 straight x plus 4 straight y right parenthesis straight m squared space equals space 4 left parenthesis straight x plus straight y right parenthesis straight m squared space space space space space space space space space space space space space space space space space

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open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets

Let C be the total cost of building tank.

therefore space space space space straight C space equals space Rs. space open square brackets 4 cross times 70 plus 4 open parentheses straight x plus 4 over straight x close parentheses cross times 45 close square brackets space equals space Rs. space open square brackets 280 plus 180 space open parentheses straight x plus 4 over straight x close parentheses close square brackets.

therefore space space dC over dx space equals space 0 plus 180 space open parentheses 1 minus 4 over straight x squared close parentheses space equals space 180 space open parentheses 1 minus 4 over straight x squared close parentheses

space dC over dx equals 0 space space space space space space space space space space space space space rightwards double arrow space space space 180 space open parentheses 1 minus 4 over straight x squared close parentheses space equals space 0 space space space space space space space rightwards double arrow space space space 1 minus 4 over straight x squared space equals 0

therefore space space space straight x squared minus 4 space equals space 0 space space space space rightwards double arrow space space straight x squared space equals space 4 space space space space space rightwards double arrow space space space straight x space equals space minus 2 comma space space 2

Rejecting x = -2 as x cannot be negative, we get x =2

fraction numerator straight d squared straight C over denominator dx squared end fraction space equals space 180 space open parentheses 0 plus 8 over straight x cubed close parentheses space equals space 1440 over straight x cubed

When

straight x space equals 2 comma space space fraction numerator straight d squared straight C over denominator dx squared end fraction space equals 1440 over 8 greater than 0

therefore space space space straight C space is space least space when space straight x space equals space 2 therefore space space space space space space space least space cost space equals space 280 plus 180 open parentheses 2 plus 4 over 2 close parentheses space equals space 280 plus 180 cross times 4 space equals space 280 plus 720 space equals space Rs. space 1000

92 Views

Short Answer Type

332.

Show that the height of a closed right circular cylinder of given volume and least surface area is equal to its diameter.

76 Views

Long Answer Type

333.

Prove that the radius of the right circular cylinder of greatest curved surface which can be inscribed in a given cone is half of that of the cone.

94 Views

Short Answer Type

334. Show that the height of the right circular cylinder of maximum volume that can be inscribed in a given circular cone of height h is

1 third straight h

82 Views

Long Answer Type

335.

Show that the height of a cylinder of maximum volume that can be inscribed in a sphere of radius R is a sphere of radius R is $fraction numerator 2 straight R over denominator square root of 3 end fraction.$ Also, find the maximum volume.

84 Views

Short Answer Type

336.

Find the height of a right circular cylinder of maximum volume, which can be inscribed in a sphere of radius 9 cm.

96 Views

337.

Show that the volume of the greatest cylinder which can be inscribed in a cone of height h and semi-vertical angle 30° is $4 over 81 πh cubed$

121 Views

Long Answer Type

338.

Show that height of the cylinder of greatest volume which can be inscribed in a right ciruclar cone of height h and semi vertical angle $straight alpha$ is one-third that of the cone and the greatest volume of cylinder is $4 over 27 πh cubed space tan squared straight alpha.$

86 Views

Short Answer Type

339. Find the volume of the largest cylinder that can be inscribed in a sphere of radius r.

75 Views

340.

Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius $5 square root of 3 space cm space is space 500 space straight pi space cm cubed.$