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Question
The volume of a closed rectangular metal box with a square base is 4096 cm3. The cost of polishing the outer surface of the box is Rs. 4 per cm2. Find the dimensions of the box for the minimum cost of polishing it.
Solution
Let the base of the box be x and height be y.
`therefore Volume = x^2y = 4096/x^2` .....(1)
∴ The total surface area is given by,
`s = 2x^2+(4x)(4096/x^2)`
∴ The cost function is given by
∴ `C(x) = 4 [2x^2 + 16384/x]Rupees` .....(2)
Differentiating w.r.t. ‘x’ we get,
`dc/dx =[4x - 16384/x^2]xx4`
Let`(dc)/(dx)=0 therefore 4x = 16384/(x^2)`
`therefore x^3 = 4096 therefore x=16`
`(d^2c)/(dx^2) at (x=16)=4xx[4+(2xx16384)/4096]= 4xx(4+8)=48>0`
Also, `y= 4096/x^2 = 4096/(16)^2= 16cm `
∴ The cost for polishing the surface area is minimum when length of base is 16 cm and height of box is 16 cm.
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