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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
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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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