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 per cm2 and the material for the sides costs Rs 2.50 per cm2. Find the least cost of the box

Answer 1

Let the length, breadth and height of the metal box be x cm, x cm and y cm respectively.

It is given that the box can contain 1024 cm³.

∴ 1024 = x²y

`=> y = 1024/x^2` .....(1)

Let C be the cost in rupees of the material used to construct.

Then

`C = 5x^2+5x^2 + 5/2 xx 4xy`

`C = 10x^2 + 10xy`

We have to find the least value of C.

`C = 10x^2 + 10xy`

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

`C = 10x^2 + 10240/x`

`=> (dC)/(dx) = 20x - 10240/x^2`

And

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

The Critical number for C are given by `(dC)/(dx) = 0`

Now

`=> (dC)/(dx) = 0`

`=> 20x - 10240/x^2 = 0`

`=> x^3 = 512`

`=> x = 8`

Also `((d^2C)/(dx^2))_(x = 8) = 20 + 20480/8^3 >0`

Thus, the cost of the box is least when x = 8.

Put x = 8 in (1), we get y = 16.

So, dimensions of the box are 8 × 8 × 16

Put x = 8, y = 16 in C = 10x² + 10xy, we get C = 1920

Hence the least cost of the box is 1920