Question Number 40479 by Raj Singh last updated on 22/Jul/18
Answered by MJS last updated on 22/Jul/18
$${a}=\mathrm{8}+\mathrm{2}{x} \\ $$$${b}={c}={d}=\mathrm{8} \\ $$$${h}=\sqrt{\mathrm{64}−{x}^{\mathrm{2}} } \\ $$$${A}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{c}\right){h}=\left(\mathrm{8}+{x}\right)\sqrt{\mathrm{64}−{x}^{\mathrm{2}} } \\ $$$${A}'\left({x}\right)=−\mathrm{2}\frac{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{32}}{\:\sqrt{\mathrm{64}−{x}^{\mathrm{2}} }} \\ $$$${A}''\left({x}\right)=\mathrm{2}\frac{{x}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{32}}{\left(\mathrm{8}−{x}\right)\sqrt{\mathrm{64}−{x}^{\mathrm{2}} }} \\ $$$${A}'\left({x}\right)=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{4} \\ $$$${A}''\left(\mathrm{4}\right)=−\mathrm{2}\sqrt{\mathrm{3}}<\mathrm{0}\:\Rightarrow\:\mathrm{maximum}\:\mathrm{at}\:{x}=\mathrm{4} \\ $$$${A}\left(\mathrm{4}\right)=\mathrm{48}\sqrt{\mathrm{3}}\approx\mathrm{83}.\mathrm{14}{cm}^{\mathrm{2}} \\ $$