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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.
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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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