Question

An open tank with a square base and vertical sides is to be constructed from a metal sheet, so as to hold a given quantity of water. Show that the total surface area is least when depth of the tank is half its width. 

Answer 1

Let the length, breadth and depth of the open tank be x, x and y, respectively. Length and breadth are same because given tank has a square base. Again, let V denotes its volume and S denotes its surface area. Now, given that

V = x²y     ......(i)

Also, we know that the total surface area of the open tank is given by

S = x² + 4xy    ....(ii)

So, depth of tank is half of its width.

or S is minimum. 

Hence, total surface area of the tank is least, when depth is half of its width

Answer 2

Let the length of  each side i.e width of square base  of the tank be  W and it's depth or vertical side be L.

Given that it holds a given quantity of water i.e it's volume remaining unaltered

It's total surface area of the open tank is

S = W^2+4WL .......(1)

Differentiating w r to W

dS/dW=2W+4L+4W(dL/dW)

For least value of surface area (dS/dW)=0

So 2W +4L+4W,(dL/dW,)=0 .......(2)

Now volume V = W^2L ......(3)

Differentiating w r to W

dV/dW=2WL+W^2(dL/dW)

But imposing the condition of definite volume we can say (dL/dW)=0.

So we get

2WL+W^2L(dL/dW)=0

=>dL/dW=- 2L/W. .....(4)

By (2) and (4) we get

2W+4L+4W(-(2L/W)=0

Hence  2W,+4L-8L=0

=> W=L/2

Width is half of depth.