A metal box with a square base and vertical sides is to contain `1024c m^3` of water, the material for the top and bottom costs Rs. `5p e rc m^2` and the material for the costs Rs. `2. 50p e rc m^2` . Find the least cost of the box.

Explanation

Since, volume of the box =1024 `cm^(3)`
Let length of the side of square base be x cm and height of the box be y cm.

`therefore` Volume of the box(V) `=x^(2).y=1024`
Since, `x^(2)y=1024 rArr y=1024/x^(2)`
Let C denotes the cost of the box.
`C=2x^(2)xx 5 +4xyxx 2.50`
`=10x^(2)(x+1024/x^(2))`
`=(10x)(x^(2))(x^(3)+1024)`
`rArr C=10x^(2)+10240/x`..............(i)
On differentiating both sides w.r.t. x, we get
`(dC)/(dx) = 20x+10240(-x)^(-2)`................(ii)
Now , `(dC)/(dx)=0`
`rArr 20x = 10240/x^(2)`
`rArr 20x^(3)=10240`
`rArr x^(3)=512 = 8^(3) rArr x=8`
Again, differentiating Eq. (ii), w.r.t. x, we get
`(d^(2)C)/(dx^(2)) = 20-10240(-2).1/x^(3)`
`=20+20480/x^(3) gt0`
`therefore (d^(2)C)/(dx^(2))_(x=8)=20+20480/512=60 gt0`
For x=8, cost is minimum and the corresponding least cost of the box,
`C(8)=10.8^(2)+10240/8`
`=640+1280=1920`
`therefore` Least cost =Rs 1920