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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 $straight x space equals 10 comma space space fraction numerator straight d squared straight C over denominator dx squared end fraction space equals space 100 plus fraction numerator 500000000 over denominator left parenthesis 10 right parenthesis to the power of 6 end fraction space equals space 100 plus 500 space equals space 600 space greater than space 0$
$therefore space space space space space straight C space is space minimum space when space straight x space equals space 10 From space left parenthesis 1 right parenthesis comma space space space straight h space equals space fraction numerator 250 over denominator left parenthesis 10 right parenthesis squared end fraction space equals space 250 over 100 space equals space 2.5 therefore space space space square space tank space is space of space side space 10 space metres space and space height space 2.5 space metres.$

441 Views

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.

87 Views

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.

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 space space space space space space space space space straight V space equals space straight x squared straight y

...(1)
Also surface area = 27 square metres

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therefore space space space from space left parenthesis 1 right parenthesis comma space space straight V space equals space straight x squared space open parentheses fraction numerator 27 minus straight x squared over denominator 4 straight x end fraction close parentheses therefore space space space space space space space space straight V space equals space 1 fourth straight x left parenthesis 27 minus straight x squared right parenthesis space equals space 1 fourth left parenthesis 27 minus straight x cubed right parenthesis space space space space space dV over dx space equals 1 fourth left parenthesis 27 minus 3 straight x squared right parenthesis 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 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 space space space space rightwards double arrow space space space space space 1 fourth left parenthesis 27 minus 3 straight x squared right parenthesis space equals 0 space space rightwards double arrow space space space 27 minus 3 straight x squared space equals 0

rightwards double arrow space space space space space straight x squared space equals space 9 space space space space space rightwards double arrow space space straight x space equals space 3 space as space straight x space cannot space be space negative space space space space space space space For space straight x space equals space 3 comma space space fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space minus 3 over 2 cross times 3 space equals space minus 9 over 2 less than 0 therefore space space space space straight V space is space maximum space when space straight x space equals space 3 therefore space space space space maximum space value space of space straight V space equals space 1 fourth left parenthesis 27 space cross times 3 space minus space 27 right parenthesis space equals space 1 fourth cross times 54 space equals space 13.5 space cubic space metres.

178 Views

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.

Let x metre be the length of a side of the removed squares. Then the height of the box is x, length is (8 – 2x) and breadth is (3 – 2 x).
Let V be the volume of the box.

therefore space space space space space straight V space space equals straight x space left parenthesis 3 minus 2 straight x right parenthesis thin space left parenthesis 8 minus 2 straight x right parenthesis space equals space 4 straight x cubed minus 22 straight x squared plus 24 straight x space space space space dV over dx space equals space and space fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space

For V to be maximum or minimum.

dV over dx space equals space 0 space space space space space space space space space space space rightwards double arrow space 12 straight x squared minus 44 straight x plus 24 space equals space 0

rightwards double arrow space space space 3 straight x squared minus 11 straight x plus 6 space equals space 0 space space space space space space space space rightwards double arrow space space space straight x space equals space fraction numerator 11 plus-or-minus square root of 121 minus 72 end root over denominator 6 end fraction rightwards double arrow space space space space space space straight x space equals space fraction numerator 11 plus-or-minus square root of 49 over denominator 6 end fraction space equals space fraction numerator 11 plus-or-minus 7 over denominator 6 end fraction space equals 3 comma space 2 over 3

Rejecting x = 3 as breadth cannot be negative, we get,

straight x space equals 2 over 3

At space space straight x space equals space 2 over 3 comma space dV over dx squared space equals space 24 space open parentheses 2 over 3 close parentheses minus 44 minus 16 minus 4 space equals space 28 less than 0

therefore space space space space space space straight V space space is space maximum space when space straight x space equals space 2 over 3 therefore space space space space space space space volume space of space largest space box space space equals open parentheses 2 over 3 close parentheses space open parentheses 3 minus 4 over 3 close parentheses space open parentheses 8 minus 4 over 3 close parentheses space equals space 2 over 3 cross times 5 over 3 cross times 20 over 3 space equals space 200 over 27 straight m cubed.

127 Views

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.

Let x (0 < x < 12) be the length of each side of the square which is to be cut from corners of the square tin sheet of each side 18 cm. Let V be the volume of the open box formed by folding up the flaps.