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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 1. x=height of cone 1−((8/x))^3 =(((x−2)/x))^3 x^3 −8^3 =(x−2)^3 x^3 −8^3 =x^3 −3×2x^2 +3×2^2 x−2^3 −6x^2 +12x+504=0 x^2 −2x−84=0 x=1+(√(85))≈10.22 cm

$${assume}\:{that}\:{the}\:{volume}\:{of}\:{cone}\:{is}\:\mathrm{1}. \\ $$$${x}={height}\:{of}\:{cone} \\ $$$$\mathrm{1}−\left(\frac{\mathrm{8}}{{x}}\right)^{\mathrm{3}} =\left(\frac{{x}−\mathrm{2}}{{x}}\right)^{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} −\mathrm{8}^{\mathrm{3}} =\left({x}−\mathrm{2}\right)^{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} −\mathrm{8}^{\mathrm{3}} ={x}^{\mathrm{3}} −\mathrm{3}×\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}×\mathrm{2}^{\mathrm{2}} {x}−\mathrm{2}^{\mathrm{3}} \\ $$$$−\mathrm{6}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{504}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{84}=\mathrm{0} \\ $$$${x}=\mathrm{1}+\sqrt{\mathrm{85}}\approx\mathrm{10}.\mathrm{22}\:{cm} \\ $$

Commented by Rasheed.Sindhi last updated on 18/Sep/19

Sir mrW Pl see Q#68761.Is it correct?

$$\boldsymbol{{Sir}}\:\boldsymbol{{mrW}}\: \\ $$$${Pl}\:{see}\:{Q}#\mathrm{68761}.{Is}\:{it}\:{correct}? \\ $$

Commented by Rasheed.Sindhi last updated on 19/Sep/19

Thanks Sir!

$$\mathbb{T}\mathrm{hanks}\:\mathbb{S}\mathrm{ir}! \\ $$

Commented by mr W last updated on 18/Sep/19

yes sir, the question is correct and good. i′ll try to solve.

$${yes}\:{sir},\:{the}\:{question}\:{is}\:{correct}\:{and}\: \\ $$$${good}.\:{i}'{ll}\:{try}\:{to}\:{solve}. \\ $$

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