A box is to build with a square base and an open top. Material for the base costs #$4/m^2#, while the material for the sides costs #$2/m^2#. Find the dimensions of the box of maximum volume which can be built at a cost of #$1200#?

1 Answer
Harish Chandra Rajpoot
Jul 19, 2018

Box of maximum volume will be a cube of edge length #10\ m#

Explanation:

Let #a# be the side of square base & #h# be the height of each of four vertical walls of the box Now, the cost of material of the box

#=4\text{(area of base)}+2\text{(area of 4 rectangular walls)}#

#=4(a^2)+2(4\cdot ah)#

#=4a^2+8ah#

But, the total cost of material of box is given as $#1200# hence we have

#4a^2+8ah=1200#

#h=\frac{300-a^2}{2a}\ ...............(1)#

Now, the volume #V# of (cuboid) box is given as

#V=text{area of square base}\times \text{height of box#

#V=a^2h#

#V=a^2(\frac{300-a^2}{2a})#

#V=150a-a^3/2#

differentiating above equation w.r.t. #a# as follows

#{dV}/{da}=d/{da}(150a-a^3/2)#

#=150-3/2a^2#

#{d^2V}/{da^2}=-3a#

For maximum volume of box, we know #{dV}/{da}=0# hence

#150-3/2a^2=0#

#a=10\ \quad (\because a>0)#

#\implies {d^2V}/{da^2}=-3(10)=-30<0#

hence the volume of box is maximum at #a=10#. Now, setting #a=10# in (1) as follows

#h=\frac{300-10^2}{2\cdot 10}#

#h=10#

Hence, the box will have square base of side #10\ m# & a vertical height of #10\ m# i.e. the box of maximum volume will be a cube of edge length #10\ m#

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