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Question Number 48879 by behi83417@gmail.com last updated on 29/Nov/18

Commented by behi83417@gmail.com last updated on 29/Nov/18

$$ABCD:squarewithsidelength=\mathbf{a}$$$$\mathbf{let}:\{\begin{array}{ll}& \mathbf{p}=EF+FG+GH+HE\\ & \mathbf{q}={EF}^2+{FG}^2+{GH}^2+{HE}^2\end{array}$$$$find:max\mathrm{\&}minof:\mathbf{p}\mathbf{\&}\mathbf{q}.$$

Commented by ajfour last updated on 29/Nov/18

Commented by mr W last updated on 29/Nov/18

$$Iguess$$$$p_{max}\Rightarrow 4a$$$$p_{min}=2\sqrt{2}a$$$$q_{max}\Rightarrow 4a^2$$$$q_{min}=2a^2$$

Commented by behi83417@gmail.com last updated on 29/Nov/18

$$q_{max}=4a^2,q_{min}=2a^2$$

Commented by behi83417@gmail.com last updated on 29/Nov/18

$$CH=z,DE=t$$$$\mathbf{q}=\mathrm{\Sigma }{EF}^2=x^2+{(a-t)}^2+y^2+{(a-x)}^2+$$$$+z^2+{(a-y)}^2+t^2+{(a-z)}^2$$$$but:x^2+{(a-x)}^2=2x^2-2ax+a^2$$$$thisismax,whenx=aandminwhen$$$$x=\frac{a}{2}.so:$$$$\frac{a^2}{2}⩽2x^2-2ax+a^2⩽a^2$$$$+\frac{a^2}{2}⩽2y^2-2ay+a^2⩽a^2\downarrow$$$$+\frac{a^2}{2}⩽2z^2-2az+a^2⩽a^2\downarrow$$$$+\frac{a^2}{2}⩽2t^2-2at+a^2⩽a^2\downarrow$$$$................................................$$$$\to 4\times \frac{a^2}{2}⩽\mathbf{q}⩽4a^2\Rightarrow 2a^2⩽\mathbf{q}⩽4a^2.$$

Commented by mr W last updated on 29/Nov/18

$$q_{max}=2\times {\left(\sqrt{2}a\right)}^2+2\times 0^2=4a^2$$$$8a^{2}wasmytypo.$$

Commented by behi83417@gmail.com last updated on 29/Nov/18

$$youarewellcomesir.$$$$sir!doyouhaveanyideafor:\mathbf{p}?$$

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