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.

Question

Filo · Accepted Answer

Let the length, width and height of the open tank be

x

,

x

and

y

with respectively. Then its volume is

x^{2} y

and total surface area is

x^{2} + 4 x y

.

It is given that the tank can be hold a given quantity of water. This mean that its volume is constant. Let it be

V

.

V = x^{2} y

The cost of the material will be least if the total surface area is least. Let's denote total surface area

S = x^{2} + 4 x y

We have to minimize

S

subjects to the condition that the volume is constant.

S = x^{2} + 4 x y

S = x^{2} + \frac{4 V}{x}

\Rightarrow \frac{d S}{d x} = 2 x - \frac{4 V}{x ^{2}}

\Rightarrow \frac{d ^{2} S}{d x ^{2}} = 2 + \frac{8 V}{x ^{3}}

For maximum or minimum value of

S

\frac{d S}{d x} = 0

\Rightarrow 2 x - \frac{4 V}{x ^{2}} = 0

\Rightarrow 2 x^{3} = 4 V

\Rightarrow 2 x^{3} = 4 x^{2} y

\Rightarrow x = 2 y

Clearly

\frac{d ^{2} x}{d x ^{2}} = 2 + \frac{8 V}{x ^{3}} > 0

for all

x

.

Here

S

is minimum when

x = 2 y

depth of

t anh

is half of width.

Question Text	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.
Answer Type	Text solution:1
Upvotes	150

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.

Text solutionVerified

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