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#?

Answer 1

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

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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Answer from HIX Tutor

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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