Question

24. 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 cost of the material will be least when depth of the tank is half of
its width
25
A
far 11
dife 1​

Answer 1

ANSWER

According to the question, the tank must hold a given quantity of water, so the volume of water can be assumed to be fixed. Let its value be denoted as C cubic units, a real constant.

The base of the tank is square, so, let the length of its side be a units and the depth of the tank be x units (a variable quantity whose value in terms of a is to be determined).

The volume of the tank is clearly area of base times depth =a2x=C(fixed).

Since the tank is open at the top, its surface area is: (bottom area + 4 walls)

a2+4ax square units.

If the cost of sheet metal material per square meter is p (fixed), then the cost of building the tank is:

p(a2+4ax)

So, the problem is to minimize p(a2+4ax) subject to the constraint a2x=C.

⇒a=xC

⇒p(a2+4ax)=p(xC+4Cx)

Differentiating the above expression with respect to x and equating to zero to find the minimum, we get:

p(−x2C+x2C)=0⇒x=34C⇒a=32C

Clearly, we can see from the above that x=2a

So, the cost of material is minimized when the depth is half of the width.