Q. 305.0( 1 Vote )

# A square tank of

Given,

Capacity of the square tank is 250 cubic metres.

Cost of the land per square meter Rs.50.

Cost of digging the whole tank is Rs. (400 × h2).

Where h is the depth of the tank.

Let us consider,

Side of the tank is x metres.

Cost of the digging is; C = 50x2 + 400h2 ---- (1)

Volume of the tank is; V = x2h ; 250 =x2h

----- (2)

Substituting (2) in (1),

----- (3)

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 ]

------- (4)

To find the critical point, we need to equate equation (4) to zero.

x6 = 106

x = 10

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:

----- (5)

[Since and]

Now let us find the value of

As , so the function C is minimum at x=10

Substituting x in equation (2)

h = 2.5 m

Therefore when the cost for the digging is minimum, when x = 10m and h = 2.5m

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