• The tank is square base open tank.
• The cost of the construction to be least.
Let us consider,
• Side of the tank is x metres.
• Height of the tank be ‘h’ metres.
• Volume of the tank is; V = x2h
• Surface Area of the tank is S = x2 +4xh
• Let Rs.P is the price per square.
Volume of the tank,
Cost of the construction be:
C = (x2 +4xh)P ---- (2)
Substituting (1) in (2),
For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function C(x) has a maximum/minimum at a point c then C’(c) = 0.
Differentiating the equation (3) with respect to x:
[Since and ]
To find the critical point, we need to equate equation (4) to zero.
x3 = 2V
Now to check if this critical point will determine the minimum volume of the tank, we need to check with second differential which needs to be positive.
Consider differentiating the equation (4) with x:
Now let us find the value of
As , so the function C is minimum at
Substituting x in equation (2)
Therefore when the cost for the digging is minimum, when and
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