constructed from a metal sheet so as to hold a given quantity of water

1.	constructed from a metal sheet so as to hold a given quantity of water. Show thatcost of the material will be least when the depth of the tank is half of itsEXAMPLE 24. An open tank with a square base and vertical sides is toptheThewidtepth
Answer» 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.

constructed from a metal sheet so as to hold a given quantity of water. Show thatcost of the material will be least when the depth of the tank is half of itsEXAMPLE 24. An open tank with a square base and vertical sides is toptheThewidtepth

Answer»

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.