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Question Number 198434 by cortano12 last updated on 20/Oct/23

Commented by mr W last updated on 20/Oct/23

$$duetosymmetry$$$$max=\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{2}}{4}$$

Answered by mr W last updated on 20/Oct/23

Commented by mr W last updated on 20/Oct/23

$$generalcase$$$$c=\sqrt{a^2+b^2}$$$$y=c-x$$$$\cos \theta =\frac{b}{\sqrt{a^2+b^2}}$$$$z=\sqrt{x^2+a^2-2ax\cos \theta }$$$$=\sqrt{x^2+a^2-\frac{2abx}{c}}$$$$P=xyz=x(c-x)\sqrt{x^2+a^2-\frac{2abx}{c}}$$$$leta=\alpha c,b=\beta c,x=\xi c$$$${\alpha }^2+{\beta }^2=1$$$$\mathrm{\Phi }=\frac{P}{c^3}=\xi (1-\xi )\sqrt{{\xi }^2+{\alpha }^2-2\alpha \beta \xi }$$$$\mathrm{\Phi }=\frac{P}{c^3}=\xi (1-\xi )\sqrt{{(\xi -\alpha \beta )}^2+{\alpha }^4}$$$$\frac{d\mathrm{\Phi }}{d\xi }=...=0$$$$...$$

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