I've solved this one by using geometry, but is there a way of finding the max by using derivatives?
My work:
So, because of geometry it has to be a cube.
The diagonal of a cube is equal to $a\sqrt{3}$ and also to $2r$
So: $a\sqrt{3}=2r$
$a=\dfrac{2r}{\sqrt{3}}$
The volume of a cube is $a^3$, so by replacing $a$ with the above result, we get:
$V=\dfrac{8r^3}{3\sqrt{3}}$, replace $r$ with $3m$ and we get $V=24\sqrt{3} \;m^3$
Any help would be appreciated!
Best Answer
For simplicity let our sphere be centred on the origin where $r=3m$, this gives us: $$f(x,y,z): x^2+y^2+z^2=r^2$$ for our constraint funtion, with $$V(B) =\prod_{i=1}^n (b_i -a_i)\to V(3)=2^3xyz$$ for the volume of a cuboid centred at the origin.
Using these we define our Lagrangian to be: $$\mathcal{L}(x,y,z)=2^3xyz-\lambda(x^2+y^2+z^2-9m^2)$$ Now solving for $\Delta\mathcal{L}(x,y,z)=0$ we obtain $$\frac{\partial\mathcal{L}}{\partial x}=0 \to4yz=\lambda x \ \ (1)$$ $$\frac{\partial\mathcal{L}}{\partial y}=0 \to4xz=\lambda y \ \ (2)$$ $$\frac{\partial\mathcal{L}}{\partial z}=0 \to4xy=\lambda z \ \ (3)$$ and our final constraint equation $x^2+y^2+z^2=9m^2$. We begin by computing $$\frac{(1)}{(2)} \to \frac{y}{x}=\frac{x}{y} \to x^2=y^2$$ Similarly, computing $$\frac{(2)}{(3)}\to y^2=z^2$$ Which then leaves us with $x^2=y^2=z^2$. Now, returning to our constraint equation we get: $$3x^2=9m^2 \ \text{which leaves us with} \ x=\sqrt{3}m$$ Thus, the maximal volume is given by $f(\sqrt{3}m,\sqrt{3}m,\sqrt{3}m)=24\sqrt{3}m^3$, thereby verifying your answer using LGM.