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Question Number 142246 by mathdanisur last updated on 28/May/21

Answered by mr W last updated on 28/May/21

$$V=\frac{1}{6}\left(\begin{array}{ccc}a& a& a\\ b& 1& b\\ a& b& b\end{array}\right)$$$$=\frac{1}{6}[a(b-b^2)-a(b^2-ab)+a(b^2-a)]$$$$=\frac{1}{6}a(b-a)(1-b)$$$$⩽\frac{1}{6}\times {\left(\frac{a+b-a+1-b}{3}\right)}^3=\frac{1}{6}\times \frac{1}{27}=\frac{1}{162}$$$$\Rightarrow V_{max}=\frac{1}{162}$$$$whena=b-a=1-b$$$$i.e.a=\frac{1}{3},b=\frac{2}{3}$$

Commented by mathdanisur last updated on 29/May/21

$$greatSirthankyou$$

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