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:The sides of the container should be 3 cm and height should be 6 cm to minimize the costStep-by-step explanation:Data provided in the question: costs  for the material used on the side = $3/cm²costs for the material used for the top lid and the base = $6/cm²Volume of the container = 54 cm³Now,let the side of the base be 'x' and the height of the box be 'y'Thus,x × x × y = 54 cm³orx²y = 54ory = [tex]\frac{54}{x^2}[/tex]  ............(1)Now,The total cost of material, CC = $3 × ( 4 side area of the box ) + $6 × (Area of the top and bottom)orC = ( $3 × 4xy ) + ( $6 × (x²)  )substituting the value of y in the above equation, we getC = [tex]3x\times4\times\frac{54}{x^2}+2\times6x^2[/tex]orC = [tex]\frac{648}{x}+12x^2[/tex]Differentiating with respect to x and putting it equals to zero to find the point of maxima of minimathus,C' = [tex]-\frac{648}{x^2}+2\times12x[/tex] = 0or[tex]2\times12x=\frac{648}{x^2}[/tex]or24x³ = 648orx = 3 cmalso,C'' = [tex]+\frac{2\times648}{x^3}+2\times12[/tex]orC''(3) = [tex]+\frac{2\times648}{3^3}+2\times12[/tex] > 0Hence,x = 3 cm is point of minimaTherefore,y = [tex]\frac{54}{x^2}[/tex]               [from 1]ory = [tex]\frac{54}{3^2}[/tex] ory = 6 cmHence,The sides of the container should be 3 cm and height should be 6 cm to minimize the cost