Question

The sum of the surface areas of cuboids with sides x, 2x and x/3 and a sphere is given to be constant. prove that the sum of their volumes is minimum if x = 3 radius of the sphere. also find the minimum value of the sum of their volumes.

Answer 1

surface area of Cuboid = 2(x.2x + 2x.x/3 + x/3.x)
= 4x² + 4x²/3 + 2x²/3 = 6x²
Surface area of sphere = 4πr² [ Let t is the radius of sphere]

Now,
sum of surface area of Cuboid + surface area of Sphere = Constant
⇒ 4πr² + 6x² = C [ where C is constant]
⇒4πr² = (C - 6x²)
⇒$\bold{r=\sqrt{\frac{C-6x^2}{4\pi}}}$----(1)

again,
volume of Cuboid = lbh = x.2x.x/3 = 2x³/3
volume of sphere = 4/3πr³
Sum of volume = V = 2x³/3 + 4/3πr³ = 2/3(x³ + 2πr³)
= $\bold{\frac{2}{3}[x^3+{2\pi(\frac{C-6x^2}{4\pi}})^{3/2}]}}$

Hence, $V=\bold{\frac{2}{3}[x^3+{2\pi(\frac{C-6x^2}{4\pi}})^{3/2}]}}$
now, differentiate with respect to x
$\bold{\frac{dV}{dx}=\frac{2}{3}[3x^2+\frac{3}{2}2\pi(\frac{C-6x^2}{4\pi})^{1/2}(-12x/4\pi)]}\\\\\bold{=\frac{2}{3}[3x^2-9xr]}$
[ From equation (1) ]
now, take dV/dx = 0 ⇒ 3x² - 9rx = 0
x = 3r is critical point , now you have to check d²V/dx² > 0 at x = 3r then, sum of volume will be minimum .
Now, again differentiate with respect to x
$\bold{\frac{d^2V}{dx^2}=\frac{2}{3}[6x -9\sqrt{\frac{C-6x^2}{4\pi}}-9x\frac{-12x/4\pi}{2\sqrt{\frac{C-6x^2}{4\pi}}}}]$
= 2/3[6x - 9r + 27/2πr ]
= 2/3 [12xr - 18r² + 27]/2πr
put x = 3r
= 2/3[ 18r²+ 27]/2πr > 0 [ because r is positive ]

Hence, d²V/dx² > 0 at x = 3r
So, the sum of volume of sphere and Cuboid is minimum at x = 3r , where r is the radius of sphere

Answer 2

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