Sphere and cone

Within the sphere of radius G = 33 cm, inscribe the cone with the largest volume. What is that volume, and what are the dimensions of the cone?

Final Answer:

r =  31.11 cm
h =  44 cm
V =  44602.24 cm³

Step-by-step explanation:

$$\begin{gathered} G=33cm \\ V=\frac{1}{3}\pi r^{2}h \\ h=G+x=33+x \\ {G}^2=x^2+r^2 \\ {r}^2=G^2-x^2 \\ V=\frac{1}{3}\pi (G^2-x^2)(G+x) \\ V=\frac{1}{3}\pi (G^3+G^{2}x-G{x}^2-x^3) \\ {V}^{\prime }=\frac{1}{3}\pi (G^2-2Gx-3x^2) \\ {V}^{\prime }=0 \\ {G}^2-2Gx-3x^2=0 \\ -3x^2-66x+1089=0 \\ 3x^2+66x-1089=0 \\ 3...\text{ prime number} \\ 66=2\cdot 3\cdot 11 \\ 1089=3^2\cdot 11^2 \\ \text{GCD}(3,66,1089)=3=3 \\ {x}^2+22x-363=0 \\ a=1;b=22;c=-363 \\ D=b^2-4ac=22^2-4\cdot 1\cdot (-363)=1936 \\ D>0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{-22\pm \sqrt{1936}}{2} \\ {x}_{1,2}=\frac{-22\pm 44}{2} \\ {x}_{1,2}=-11\pm 22 \\ {x}_1=11 \\ {x}_2=-33 \\ \text{ Factored form of the equation: } \\ (x-11)(x+33)=0 \\ h=G+x_1=33+11=44cm \\ r=\sqrt{G^2-x_1^2}=31.11\text{cm} \end{gathered}$$

$h=33+11=44\text{cm}$

$V=\frac{1}{3}\pi r^{2}h=44602.24{\text{cm}}^3$

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