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 material will be least when the depth of the tank is half of its width. If the cost is to be least when depth of the tank is half of its width. If the cost is to be borne by nearby settled lower income families, for whom water will be provided, what kind of value is hidden in this question?

Solution

Let the length, width and height of the open tank be x, x and y units respectively. Then, its volume is x²Y and the total surface area is x² + 4xy.

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

V = x²y

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

Then,

S = x² + 4xy

We have to minimize S object to the condition that the volume V is constant.

Now,

$S = x^{2} + 4 xy \Rightarrow S = x^{2} + \frac{4 V}{x} \Rightarrow \frac{dS}{dx} = 2 x - \frac{4 V}{x^{2}} and \frac{d^{2} S}{{dx}^{2}} = 2 + \frac{8 V}{x^{3}}$

The critical numbers of S are given by dS/dx = 0

Now, ds/dx = 0

$⇒2x - \frac{4 V}{x^{2}} = 0 \Rightarrow 2 x^{3} - 4 V = 0 2 x^{3} = 4 x^{2} y x = 2 y Clearly, \frac{d^{2} S}{{dx}^{2}} = 2 + \frac{8 V}{x^{3}} > 0 for all x$

Hence, S is minimum when x =2y i.e the depth (height) of the tank is half of its width.

For Daily Free Study Material Join wiredfaculty

Application Of Derivatives

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction