Question

34. The sum of the surface areas of a rectangular parallelepiped with sides x, 2x and x/3
and a sphere is given to be
constant. Prove that the sum of their volumes is minimum, ifx is equal to three times the radius of the sphere.
find the minimum value of the sum of their volumes.​

Answer 1

Answer:

The minimum value of sum of volume of parallelepiped and sphere is $\dfrac{(54+4\pi )r^{3}}{3}$

Step-by-step explanation:

Given as :

The three side of rectangular parallelepiped are

Length = x unit

width = 2 x unit

Height = $\dfrac{x}{3}$ unit

So, Surface area of parallelepiped = 2 (length × width + width × height + height × length)

Or, Surface area of parallelepiped = 2 (x unit × 2 x unit + 2 x unit ×$\dfrac{x}{3}$  unit + $\dfrac{x}{3}$ unit × x unit)

Or, Surface area of parallelepiped = 2 ( $\dfrac{9x^{2} }{3}$ )

Or, Surface area of parallelepiped = 2 × 3 x²

∴ Surface area of parallelepiped = 6 x²

And Volume of parallelepiped = v = length × width × height

Or, v = x × 2 x × $\dfrac{x}{3}$

i.e  v = $\frac{2x^{3}}{3}$

Again

Let The radius of sphere = r unit

So, Surface area of sphere = 4 × π × radius²

i.e Surface area of sphere = 4 π r²

Volume of sphere = $\dfrac{4}{3}$ × π × radius³

Or, V = $\dfrac{4}{3}$ × π × r³

i.e V = $\dfrac{4\pi r^{3}}{3}$

So, The of Surface area of parallelepiped and  Surface area of sphere = constant

i.e 6 x² + 4 π r² = constant

Again

For, x = 3 r

volume of parallelepiped + volume of sphere = $\dfrac{2x^{3}}{3}$ + $\dfrac{4\pi r^{3}}{3}$

or, volume of parallelepiped + volume of sphere = $\dfrac{54(r)^{3}}{3}$ + $\dfrac{4\pi r^{3}}{3}$

or ,  volume of parallelepiped + volume of sphere = $\dfrac{(54+4\pi )r^{3}}{3}$

Hence, The minimum value of sum of volume of parallelepiped and sphere is $\dfrac{(54+4\pi )r^{3}}{3}$  Answer