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Question Number 131960 by liberty last updated on 10/Feb/21

Answered by mr W last updated on 10/Feb/21

$$r^2+h^2={\left(\sqrt{3}\right)}^2\Rightarrow r^2=3-h^2$$$$V=\frac{\pi }{3}{r}^{2}h=\frac{\pi }{3}(3-h^2)h$$$$\frac{dV}{dh}=\frac{\pi }{3}(3-3h^2)=0$$$$\Rightarrow h^2=1$$$$\Rightarrow h=1$$$$\Rightarrow r=\sqrt{3-1}=\sqrt{2}$$$$\frac{d^{2}V}{{dh}^2}=\frac{\pi }{3}(-6h)<0\Rightarrow maximum$$$$\Rightarrow V_{max}=\frac{\pi }{3}\times {\left(\sqrt{2}\right)}^2\times 1=\frac{2\pi }{3}$$

Commented by otchereabdullai@gmail.com last updated on 10/Feb/21

$$\mathrm{thank}\mathrm{you}\mathrm{prorW}$$

Commented by liberty last updated on 10/Feb/21

$$\mathrm{nice}$$

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