Question

For a rectangular cube with a square base, suppose that it costs $3/cm^2 for the material used on the side and $6/cm^2 for the material used for the top lid and the base. Assuming that the volume of this container is 54 cm^3 , what is the side length of the cube with the smallest cost?

Answer 1

Answer:

The sides of the container should be 3 cm and height should be 6 cm to minimize the cost

Step-by-step explanation:

Data provided in the question:

costs  for the material used on the side = $3/cm²

costs for the material used for the top lid and the base = $6/cm²

Volume of the container = 54 cm³

Now,

let the side of the base be 'x' and the height of the box be 'y'

Thus,

x × x × y = 54 cm³

or

x²y = 54

or

y = $\frac{54}{x^2}$  ............(1)

Now,

The total cost of material, C

C = $3 × ( 4 side area of the box ) + $6 × (Area of the top and bottom)

or

C = ( $3 × 4xy ) + ( $6 × (x²)  )

substituting the value of y in the above equation, we get

C = $3x\times4\times\frac{54}{x^2}+2\times6x^2$

or

C = $\frac{648}{x}+12x^2$

Differentiating with respect to x and putting it equals to zero to find the point of maxima of minima

thus,

C' = $-\frac{648}{x^2}+2\times12x$ = 0

or

$2\times12x=\frac{648}{x^2}$

or

24x³ = 648

or

x = 3 cm

also,

C'' = $+\frac{2\times648}{x^3}+2\times12$

or

C''(3) = $+\frac{2\times648}{3^3}+2\times12$ > 0

Hence,

x = 3 cm is point of minima

Therefore,

y = $\frac{54}{x^2}$               [from 1]

or

y = $\frac{54}{3^2}$

or

y = 6 cm

Hence,

The sides of the container should be 3 cm and height should be 6 cm to minimize the cost