A closed rectangular storage bin is to be made so that it has a square base. The volume of the bin must be 8m^3. The material to make the sides costs twice as much as that for the top and the bottom. Find the dimensions of the box that will (- cont.)?

Let the bin have square base of length #x# metres and a height of #y# metres.

Then the volume of the bin would equal #8#=#x^2y#.......#[1]#
The surface area of the box will be #2x^2#[ the top and bottom squares]+#4xy#, the area of the four sides.

Therefore total Area, #A #=#2x^2+4xy#, but the cost of the two bases is only half the cost per metre squared of the sides of the bin so we can say, Cost #C# = #x^2+4xy#........#[2]#

From ....#[1]# #y=8/x^2# and substituting this value for #y# in .....#[2]#
#C=x^2+4x[8/x^2]# = #x^2+32/x#.........#[3]#

Differentiating ........#[3]# with respect to #x# gives #[2x-16/x^2]# and setting this equal to zero for stationary turning point[ max/min]

#x^3=16, i.e, #x# #=#root3 16# = #+-2.5198# [ approx, and taking the second derivative] #d^2y/dx^2# is positive when #x=2.5198# and must therefore be the dimension of #x# that minimises the cost.Substituting this value for #x# in #y =8/x^2# will yield the required value of #y#, as in answer above.

Hope this helps.

Let us set up the following variables:

# { (l, "length (or width) of square base (m)"), (h, "Height of storage bin (m)"), (C, "Total cost") :} #

We can calculate the total volume of the box using #w xx l xx h# and we know that this volume is fixed, thus:

# l xx l xx h = 8=> l^2h=8 # ..... [A]

Let us also assume that the cost per square metre for the top or bottom) is #n# per #m^2# (in some convenient unit of currency). Given that cost is not further referenced in the question, we expect this variable to be eliminated.

Given this, we can calculate the surface area of the identical top and bottom, and therefore the associated cost:

# \ \ \ \ \A_("top/bottom") = 2 xx l xx l = 2l^2#

# :. C_("top/bottom") = 2nl^2 #

Similarly, we can calculate the surface area of the identical sides, and therefore the associated cost (which is twice that of the top & bottom):

# \ \ \ \ \A_("sides") = 4 xx l xx h = 4lh#

# :. C_("sides") = (2n)(4lh) =8nlh #

Thus, we can form an expression for the total cost:

# C = 2nl^2 + 8nlh #

Using the result [A] we can write #h=8/l^2# and eliminate #h#:

# C = 2nl^2 + 8nl(8/l^2) #
# \ \ = 2nl^2 + (64n)/l #

We now have the cost function #C# define as a function of a single variable #l#, so we can now utilise calculus in an attempt to minimize the cost, so differentiating wrt #l#, we have:

# (dC)/(dl) = 4nl - (64n)/l^2 #

We seek a critical point which requires that the first derivative vanish, so we require:

# (dC)/(dl) = 0 => 4nl - 64n/l^2 #

# :. (4n(l^3 - 16))/l^2 = 0 #

# :. l^3 - 16 = 0 #

# :. l = root(3)(16) #

With this value of #l# we have using [A] that:

# h=8/l^2 = 8/(root(3)(16))^2 #

Thus, the sought dimensions (to 2dp as indicated) are:

# l = 2.52# and #h=1.26 \ (m)#

We should check that these value leads to a minimum (rather than a maximum) cost. As the size of the box is finite this should really be intuitive. And we can confirm this graphically:

graph{y=2x^2 + 64/x [-20, 20, -150, 150]}