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A metal box with a square base and vertical sides is to contain 1024 cm3. The material for the top and bottom costs Rs 5/cm2 and the material for the sides costs Rs 2.50/cm2. Find the least cost

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Question

A metal box with a square base and vertical sides is to contain 1024 cm3. The material for the top and bottom costs Rs 5/cm2 and the material for the sides costs Rs 2.50/cm2. Find the least cost of the box.

Sum
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Solution


Let x be the side of the square base and y be the length of the vertical sides.

Area of the base and bottom = 2x2 cm2

∴ Cost of the material required = ₹ 5 × 2x2

= ₹ 10x2

Area of the 4 sides = 4xy cm2

∴ Cost of the material for the four sides

= ₹ 2.50 x 4xy

= ₹ 10xy

Total cost C = 10x2 + 10xy  .....(i)

New volume of the box = x × x × y

⇒ 1024 = x2y

∴ y = `1024/x^2`  ....(ii)

Putting the value of y in equation (i) we get

C = `10x^2 + 10x xx 1024/x^2`

⇒ C = `10x^2 + 10240/x`

Differentiating both sides w.r.t. x, we get

`"dC"/"dx" = 20x - 10240/x^2`  ....(iii)

For local maxima and local minima `"dC"/"dx"` = 0

`20 - 102400/x^2` = 0

⇒ 20x3 – 10240 = 0

⇒ x3 = 512

⇒ x = 8 cm

Now from equation (ii)

y = `10240/(8)^2`

= `10240/64`

= 16 cm

∴ Cost of material used C = 10x2 + 10xy

= 10 × 8 × 8 + 10 × 8 × 16

= 640 + 1280

= 1920

Now differentiating equation (iii) we get

`("d"^2"C")/("dx"^2) = 20 + 20480/x^3`

Put x = 8

= `20 + 20480/(8)^3`

= `20 + 20480/512`

= 20 + 40 = 60 > 0 minima

Hence, the required cost is ₹ 1920 which is the minimum.

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Chapter 6: Application Of Derivatives - Exercise [Page 138]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 6 Application Of Derivatives
Exercise | Q 33 | Page 138

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