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. 

Answer 1

 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.