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Solution for An Open Tank is to Be Constructed with a Square Base and Vertical Sides So as to Contain a Given Quantity of Water. Lead with Be Least, If Depth is Made Half - CBSE (Commerce) Class 12 - Mathematics

Question

An open tank is to be constructed with a square base and vertical sides so as to contain a given quantity of water. Show that the expenses of lining with lead with be least, if depth is made half of width.

Solution

$\text { Let l, h, V and S be the length, height, volume and surface area of the tank to be constructed }.$

$\text { Since volume, V is constant,}$

$l^2 h = V$

$\Rightarrow h = \frac{V}{l^2} . . . \left( 1 \right)$

$\text { Surface area, S = } l^2 + 4lh$

$\Rightarrow S = l^2 + \frac{4V}{l} \left[\text { From eq } . \left( 1 \right) \right]$

$\Rightarrow \frac{dS}{dl} = 2l - \frac{4V}{l^2}$

$\text { For S to be maximum or minimum, we must have }$

$\frac{dS}{dl} = 0$

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

$\Rightarrow 2 l^3 - 4V = 0$

$\Rightarrow 2 l^3 = 4V$

$\Rightarrow l^3 = 2V$

$\text { Now, }$

$\frac{d^2 S}{d l^2} = 2 + \frac{8V}{l^3}$

$\Rightarrow \frac{d^2 S}{d l^2} = 2 + \frac{8V}{2V} = 6 > 0$

$\text { Here, surface area is minimum }.$

$h = \frac{V}{l^2}$

$\text { Substituting the value of V } = \frac{l^3}{2}\text { in eq } . \left( 1 \right),\text { we get }$

$h = \frac{l^3}{2 l^2}$

$\Rightarrow h = \frac{l}{2}$

$\text { Hence proved }.$

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Solution An Open Tank is to Be Constructed with a Square Base and Vertical Sides So as to Contain a Given Quantity of Water. Lead with Be Least, If Depth is Made Half Concept: Graph of Maxima and Minima.
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