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Class 12th Application of Derivatives

The sum of the surface areas of a rectangular parallelepiped with sides x, 2x, and and a sphere is given to be constant. Prove that the sum of their volumes is minimum if x is equal to three times the radius of the sphere. Also, find the minimum value of the sum of their volumes.

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t' r' b e t h e r a d i u s o f t h e s p h e r e . \\ ∴ S u r f a c e a r e a o f t h e s p h e r e = 4 π r^{2} \\ V o l u m e o f t h e s p h e r e = \frac{4}{3} π r^{3} \\ T h e s i d e s o f t h e p a r a l l e l o p i p e d a r e x, 2 x a n d \frac{x}{3} \\ ∴ I t s s u r f a c e a r e a = 2 [x * 2 x + 2 x * \frac{x}{3} + x * \frac{x}{3}] \\ = 2 [2 x^{2} + \frac{2 x^{2}}{3} + \frac{x^{2}}{3}] = 2 [2 x^{2} + x^{2}] \\ = 2 [3 x^{2}] = 6 x^{2} \\ V o l u m e o f t h e p a r a l l e l o p i p e d = x * 2 x * \frac{x}{3} = \frac{2}{3} x^{3} \\ A s p e r t h e c o n d i t i o n s o f t h e q u e s t i o n, \\ s u r f a c e a r e a o f t h e p a r a l l e l o p i p e d + s u r f a c e a r e a o f t h e s p h e r e = constant \\ \Rightarrow 6 x^{2} + 4 π r^{2} = K (constant) \Rightarrow 4 π r^{2} = K - 6 x^{2} \\ ∴ r^{2} = \frac{K - 6 x^{2}}{4 π} \dots (i) \\ N o w l e t V = V o l u m e o f t h e p a r a l l e l o p i p e d + V o l u m e o f t h e s p h e r e \\ \Rightarrow V = \frac{2}{3} x^{3} + \frac{4}{3} π r^{3} \\ \Rightarrow V = \frac{2}{3} x^{3} + \frac{4}{3} π {[\frac{K - 6 x^{2}}{4 π}]}^{3 / 2} [F r o m e q . (i)] \\ \Rightarrow &thins \end{array}$

$\begin{array}{l} S q u a r i n g b o t h s i d e s, w e g e t \\ \Rightarrow 4 π x^{2} = 9 (K - 6 x^{2}) \Rightarrow 4 π x^{2} = 9 K - 54 x^{2} \\ \Rightarrow 4 π x^{2} + 54 x^{2} = 9 K \\ \Rightarrow K = \frac{4 π x^{2} + 54 x^{2}}{9} (i i) \\ \Rightarrow 2 x^{2} (2 π + 27) = 9 K \\ ∴ x^{2} = \frac{9 K}{2 (2 π + 27)} = 3 \\ N o w f r o m e q . (i) w e h a v e \\ r^{2} = \frac{K - 6 x^{2}}{4 π} \\ \Rightarrow r^{2} = \frac{\frac{4 π x^{2} + 54 x^{2}}{9} - 6 x^{2}}{4 π} \\ \Rightarrow r^{2} = \frac{4 π x^{2} + 54 x^{2} - 54 x^{2}}{9 * 4 π} = \frac{4 π x^{2}}{9 * 4 π} = \frac{x^{2}}{9} \Rightarrow r = \frac{x}{3} ∴ x = 3 r \\ N o w w e h a v e \frac{d V}{d x} = 2 x^{2} - \frac{3 x}{} {(K - 6 x^{2})}^{1 / 2} \end{array}$

$\begin{array}{l} = 12 - \frac{3}{} \frac{\frac{4 K π + 54 K - 108 K}{4 π + 54}}{} \\ = 12 - \frac{3}{} [\frac{\frac{4 K π - 54 K}{4 π + 54}}{}] \\ = 12 - \frac{3}{} [\frac{4 K π - 54 K}{.}] \\ = 12 - \frac{6}{} [\frac{2 π - 27}{.}] \\ = 12 + \frac{6 K}{} [\frac{27 - 2 π}{.}] > 0 [? 27 - 2 π > 0] \\ ∴ \frac{d^{2} V}{d x^{2}} > 0 S o, i t i s minima. \\ H e n c e, t h e s u m o f v o l u m e i s min \end{array}$

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Class 12th Application of Derivatives

A metal box with a square base and vertical sides is to contain 1024 cm³. The material for the top and bottom costs Rs 5/cm² and the material for the sides costs Rs 2.50/cm². Find the least cost of the box

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Question as classified in NCERT Exemplar

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Class 12th Application of Derivatives

AB is a diameter of a circle and C is any point on the circle. Show that the area of ?ABC is maximum when it is isosceles.

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

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Class 12th Application of Derivatives

AB is a diameter of a circle and C is any point on the circle. Show that the area of ?ABC is maximum when it is isosceles.

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New answer posted

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Class 12th NEET

Consider an electron in front of metallic surface of a distance d. Assume the force of attraction by the plate is given as $\frac{1}{4} \frac{q^{2}}{4 π ε_{0} d^{2}}$ . Calculate work in taking the to an infinite distance from the plate .Taking d=0.1 nm. find the work done in electron volts?

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Pallavi Pathak

Contributor-Level 10

Explanation- F_x= $\frac{1}{4} \frac{q^{2}}{4 ? ?_{0} x^{2}}$

W= $?_{d}^{?} f d x$ = $?_{d}^{?} \frac{q^{2} d x}{4 * 4 ? ?_{0}} \frac{1}{x^{2}}$

= $\frac{q^{2}}{4 * 4 ? ?_{0}} \frac{1}{d}$

= $\frac{{(1.6 * 10^{- 19})}^{2} * 9 * 10^{9}}{4 * 10^{- 10}}$ J= 3.6eV

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Class 12th Application of Derivatives

If the sum of the surface areas of a cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} N o w p u t t i n g t h e v a l u e o f K i n e q . (i) w e g e t \\ 6 x^{2} + 4 π r^{2} = x^{2} (π + 6) \\ \Rightarrow 6 x^{2} + 4 π r^{2} = π x^{2} + 6 x^{2} \Rightarrow 4 π r^{2} = π x^{2} \Rightarrow 4 r^{2} = x^{2} \\ ∴ 2 r = x \\ ∴ x : 2 r = 1 : 1 \\ N o w d i f f e r e n t i a t i n g e q . (i i) w . r . t ., x, w e h a v e \\ \frac{d^{2} V}{d x^{2}} = 6 x - \frac{3}{} \frac{d}{d x} [x {(K - 6 x^{2})}^{1 / 2}] \\ = 6 x - \frac{3}{} [x . \frac{1}{2} * (- 12 x) + {(K - 6 x^{2})}^{1 / 2} . 1] \\ = 6 x - \frac{3}{} [\frac{- 6 x^{2}}{} +] \\ = 6 x - \frac{3}{} [\frac{- 6 x^{2} + K - 6 x^{2}}{}] = 6 x + \frac{3}{} [\frac{12 x^{2} - K}{}] \\ = 6 + \frac{3}{} [\frac{\frac{12 K}{π + 6} - K}{}] [P u t x =] \\ = 6 + \frac{3}{} [\frac{12 K - π K - 6 K}{}] = 6 + \frac{3}{} [\frac{6 K - π K}{}] \\ = 6 + \frac{3}{π} [(6 K - π K)] > 0 s o i t i s minima . \\ H e n c e, t h e r e q u i r e d r a t i o i s 1 : 1 w h e n t h e c o m b i n e d v o l u m e i s minimum . \end{array}$

Minimum

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New question posted

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Class 12th Application of Derivatives

If the sum of the surface areas of a cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?

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New answer posted

a year ago

Class 12th Application of Derivatives

Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible when revolved about one of its sides. Also, find the maximum volume.

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

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Class 12th NCERT

Consider a thin target (10^–2m square, 10^–3m thickness) of sodium, which produces a photocurrent of 100μA when a light of intensity 100W/m²(λ = 660nm) falls on it. Find the probability that a photoelectron is produced when a photons strikes a sodium atom. [Take density of Na = 0.97 kg/m³]

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Pallavi Pathak

Contributor-Level 10

Explanation- A= 10^-4m²

So d= 10^-3 and i= 100^-4A

I= 100W/m²

$? = 600 n m = 600 * 10^{- 9}$

$?_{N a} = 0.97 k g / m^{3}$

$v o l u m e = A * d$ 10^-4(10^-3)=10^-7m³

$f o r 23 k g o f s o i d u m$

So volume Na atoms=23/0.97m³

Volume occupied by one Na atom= $\frac{23}{0.97 * 6 * 10^{26}} = 3.95 * 10^{- 26} m^{3}$

Number of Na atoms in target $\frac{10^{- 7}}{3.95 * 10^{- 26}} = 2.53 * 10^{18}$

So energy falling per sec= $\frac{n h c}{?} = I A$

So n= $\frac{I A ?}{h c}$ = $\frac{100 * 10^{- 4} * 660 * 10^{- 9}}{6.62 * 10^{- 34} * 3 * 10^{8}} = 3.3 * 10^{16}$

N=P $* n * N a = P * 3.3 * 10^{16} * 2.53 * 10^{18}$

I = 100 $* 10^{- 6} = 10^{- 4}$ A

I=Ne= $P * 3.3 * 10^{16} * 2.53 * 10^{18}$ ( $10^{- 4}$ A)

P= 7.48 $* 10^{- 21}$ it is less than 1.

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Class 12th Application of Derivatives

An open box with a square base is to be made of a given quantity of cardboard of area c². Show that the maximum volume of the box is $\frac{c ³}{6 \sqrt 3}$ cubic units.