# Question-68962

Question Number 68962 by ahmadshah last updated on 17/Sep/19
Commented by mr W last updated on 17/Sep/19
$${assume}\:{that}\:{the}\:{volume}\:{of}\:{cone}\:{is}\:\mathrm{1}. \\$$$${x}={height}\:{of}\:{cone} \\$$$$\mathrm{1}−\left(\frac{\mathrm{8}}{{x}}\right)^{\mathrm{3}} =\left(\frac{{x}−\mathrm{2}}{{x}}\right)^{\mathrm{3}} \\$$$${x}^{\mathrm{3}} −\mathrm{8}^{\mathrm{3}} =\left({x}−\mathrm{2}\right)^{\mathrm{3}} \\$$$${x}^{\mathrm{3}} −\mathrm{8}^{\mathrm{3}} ={x}^{\mathrm{3}} −\mathrm{3}×\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}×\mathrm{2}^{\mathrm{2}} {x}−\mathrm{2}^{\mathrm{3}} \\$$$$−\mathrm{6}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{504}=\mathrm{0} \\$$$${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{84}=\mathrm{0} \\$$$${x}=\mathrm{1}+\sqrt{\mathrm{85}}\approx\mathrm{10}.\mathrm{22}\:{cm} \\$$
Commented by Rasheed.Sindhi last updated on 18/Sep/19
$$\boldsymbol{{Sir}}\:\boldsymbol{{mrW}}\: \\$$$${Pl}\:{see}\:{Q}#\mathrm{68761}.{Is}\:{it}\:{correct}? \\$$
Commented by Rasheed.Sindhi last updated on 19/Sep/19
$$\mathbb{T}\mathrm{hanks}\:\mathbb{S}\mathrm{ir}! \\$$
Commented by mr W last updated on 18/Sep/19
$${yes}\:{sir},\:{the}\:{question}\:{is}\:{correct}\:{and}\: \\$$$${good}.\:{i}'{ll}\:{try}\:{to}\:{solve}. \\$$