Question Number 71946 by oyemi kemewari last updated on 22/Oct/19
Answered by MJS last updated on 23/Oct/19
$$\mathrm{12}=\mathrm{3}{y}\left(\frac{\mathrm{3}{y}}{\mathrm{3}+\mathrm{3}{y}}\right)^{\mathrm{2}/\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{400}}\right)^{\mathrm{1}/\mathrm{2}} \\ $$$$\mathrm{80}={y}\left(\frac{{y}}{{y}+\mathrm{1}}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\mathrm{512000}={y}^{\mathrm{3}} \frac{{y}^{\mathrm{2}} }{\left({y}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${y}^{\mathrm{5}} −\mathrm{512000}{y}^{\mathrm{2}} −\mathrm{1024000}{y}−\mathrm{512000}=\mathrm{0} \\ $$$$\Rightarrow\:{y}\approx\mathrm{80}.\mathrm{6599} \\ $$
Commented by oyemi kemewari last updated on 25/Oct/19
thanks sir
Commented by oyemi kemewari last updated on 25/Oct/19
but how did you get y from line 4
Commented by MJS last updated on 26/Oct/19
$$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$