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Question-40479




Question Number 40479 by Raj Singh last updated on 22/Jul/18
Answered by MJS last updated on 22/Jul/18
a=8+2x  b=c=d=8  h=(√(64−x^2 ))  A(x)=(1/2)(a+c)h=(8+x)(√(64−x^2 ))  A′(x)=−2((x^2 +4x−32)/( (√(64−x^2 ))))  A′′(x)=2((x^2 −8x−32)/((8−x)(√(64−x^2 ))))  A′(x)=0 ⇒ x=4  A′′(4)=−2(√3)<0 ⇒ maximum at x=4  A(4)=48(√3)≈83.14cm^2
$${a}=\mathrm{8}+\mathrm{2}{x} \\ $$$${b}={c}={d}=\mathrm{8} \\ $$$${h}=\sqrt{\mathrm{64}−{x}^{\mathrm{2}} } \\ $$$${A}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{c}\right){h}=\left(\mathrm{8}+{x}\right)\sqrt{\mathrm{64}−{x}^{\mathrm{2}} } \\ $$$${A}'\left({x}\right)=−\mathrm{2}\frac{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{32}}{\:\sqrt{\mathrm{64}−{x}^{\mathrm{2}} }} \\ $$$${A}''\left({x}\right)=\mathrm{2}\frac{{x}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{32}}{\left(\mathrm{8}−{x}\right)\sqrt{\mathrm{64}−{x}^{\mathrm{2}} }} \\ $$$${A}'\left({x}\right)=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{4} \\ $$$${A}''\left(\mathrm{4}\right)=−\mathrm{2}\sqrt{\mathrm{3}}<\mathrm{0}\:\Rightarrow\:\mathrm{maximum}\:\mathrm{at}\:{x}=\mathrm{4} \\ $$$${A}\left(\mathrm{4}\right)=\mathrm{48}\sqrt{\mathrm{3}}\approx\mathrm{83}.\mathrm{14}{cm}^{\mathrm{2}} \\ $$

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