An open rectangular tank has a horizontal square base of side x metres. The height of
the tank is y metres. The tanks is made of thin sheet metal. The external surface area of
the tank is S square metres and the volume is V cubic metres.
First of all we need a good diagram of the tank and its net:
From the diagram we can work out equations for the surface area and the volume.
The surface area is one square (x by x) and four rectangles (y by x)
So S = x2 + 4xy . . . . . (1)
The volume is given by base area (x2) times height (y)
So V = x2y . . . . . (2)
The equation we are asked to prove contains S, V and x but not y,
so use equations (1) and (2) above to eliminate y as follows:
which is the equation required.
Now replace V by the given value 62.5 and differentiate with respect to x:
The maximum or minimum value of S is given when dS/dx = 0
So the base is a square of side 5 metres. The height y can be found by replacing
x by 5 in the equation used for eliminating y above:
Finally we need to check whether the surface area is a maximum or a minimum when x=5 and y=2.5
This can be done by considering the gradient either side of the point when x=5
We draw up a table:
Therefore the tank which meets the stated requirements with a minimum surface area has base 5 metres by 5 metres and height 2.5 metres.
We may wish to check this answer. Plot the graph of the surface area
S = 250/x + x2
The minimum point does indeed seem to be where x=5, and this can be confirmed by zooming in.