Minimum Volume Enclosed

Let the surface area of the sphere be $x$ then that of the cube is $10-x$ . Then $x = 4\pi r^2$ , where $r$ is the radius of the sphere. $\implies r = \sqrt{\dfrac x{4\pi}}$ . The volume of the sphere $V_s = \dfrac 43 \pi r^3 = \dfrac {4\pi}3 \left(\dfrac x{4\pi}\right)^\frac 32 = \dfrac {x^\frac 32}{3\sqrt{4\pi}}$ .

The surface area of the cube $(10-x) = 6a^2$ , where $a$ is the side length of the cube. $\implies a = \sqrt{\dfrac {10-x}6}$ . Then the volume of the cube $V_c = a^3 = \left(\dfrac {10-x}6\right)^\frac 32$ .

The combined volume is therefore,

$\begin{aligned} V & = V_s+V_c \\ & = \frac {x^\frac 32}{3\sqrt{4\pi}} + \left(\frac {10-x}6\right)^\frac 32 \\ \frac {dV}{dx} & = \frac {x^\frac 12}{2\sqrt{4\pi}} - \frac 14\left(\frac {10-x}6\right)^\frac 12 & \small {\color{#3D99F6}\text{Putting }\frac {dV}{dx} = 0} \\ \frac {x^\frac 12}{2\sqrt{4\pi}} & = \frac 14\left(\frac {10-x}6\right)^\frac 12 \\ \frac {x}{16\pi} & = \frac {10-x}{16\cdot 6} \\ 6x & =10 \pi - \pi x \\ \implies x & = \frac {10\pi}{6+\pi} \end{aligned}$

Note that $\dfrac {d^2V}{dx^2} = \dfrac 1{4\sqrt{4\pi x}} + \dfrac 1{48} \sqrt{\dfrac 6{10-x}} > 0$ . This means that $V$ is minimum when $x = \dfrac {10\pi}{6+\pi}$ and:

$\begin{aligned} V_{min} & = \frac {\left(\frac {10\pi}{6+\pi}\right)^\frac 32}{3\sqrt{4\pi}} + \left(\frac {10-\frac \pi{6+\pi}}6\right)^\frac 32 \approx \boxed{1.743} \end{aligned}$

Lagrangian method of multipliers is the shortest method for this optimisation problem realising: 10=4πx^2+6y^2=g(x,y)

4/3πx^3+y^3=f(x,y)=V

Now div{f(x,y)}=¥div{g(x,y)}

Where ¥ is the constant to be determined which comes up to ¥=(10/(96+16π))^(3/2)

And hence the ans is V=¥*((32π/3)+64))

which is

[(10/(96+16π))^(3/2)]*((32π/3)+64))=the given ans