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