Question-223424

Question Number 223424 by behi834171 last updated on 25/Jul/25

Commented by fantastic last updated on 25/Jul/25

$${GIGANTIC}\:{when}\:{using}\:{normal}\:{method} \\ $$

Commented by Ghisom last updated on 25/Jul/25

p_(min) =4(√5) at x=((317−32(√(61)))/(625))

$${p}_{\mathrm{min}} =\mathrm{4}\sqrt{\mathrm{5}}\:\mathrm{at}\:{x}=\frac{\mathrm{317}−\mathrm{32}\sqrt{\mathrm{61}}}{\mathrm{625}} \\ $$

Commented by fantastic last updated on 25/Jul/25

$${how} \\ $$

Answered by behi834171 last updated on 25/Jul/25

x≥0 ,x≤1⇒x∈[0,1] 1)41−40(√x)=(16+25)−2×4×5(√x)= =4^2 +5^2 −2×4×5×cos𝛉 (cosinus rule) 2)89−80(√(1−x))=(25+64)−2×5×8(√(1−x))= =5^2 +8^2 −2×5×8×sin𝛉= =5^2 +8^2 −2×5×8×cos((𝛑/2)−𝛉) 3)[see the pic below⇓⇓] p,is minimum,when the points: B and D,C,are colinear. so,because of the triangle:ABC, is rightangle,then: BC=p_(min) =(√(4^2 +8^2 ))=4(√5) .■

$$\boldsymbol{{x}}\geqslant\mathrm{0}\:\:,\boldsymbol{{x}}\leqslant\mathrm{1}\Rightarrow\boldsymbol{{x}}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\mathrm{41}−\mathrm{40}\sqrt{\boldsymbol{{x}}}=\left(\mathrm{16}+\mathrm{25}\right)−\mathrm{2}×\mathrm{4}×\mathrm{5}\sqrt{\boldsymbol{{x}}}= \\ $$$$=\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{2}×\mathrm{4}×\mathrm{5}×\boldsymbol{{cos}\theta}\:\:\left({cosinus}\:{rule}\right) \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\mathrm{89}−\mathrm{80}\sqrt{\mathrm{1}−\boldsymbol{{x}}}=\left(\mathrm{25}+\mathrm{64}\right)−\mathrm{2}×\mathrm{5}×\mathrm{8}\sqrt{\mathrm{1}−\boldsymbol{{x}}}= \\ $$$$=\mathrm{5}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} −\mathrm{2}×\mathrm{5}×\mathrm{8}×\boldsymbol{{sin}\theta}= \\ $$$$=\mathrm{5}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} −\mathrm{2}×\mathrm{5}×\mathrm{8}×\boldsymbol{{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{\theta}\right) \\ $$$$ \\ $$$$\left.\mathrm{3}\right)\left[\boldsymbol{{see}}\:\boldsymbol{{the}}\:\boldsymbol{{pic}}\:\boldsymbol{{below}}\Downarrow\Downarrow\right] \\ $$$$ \\ $$$$\boldsymbol{{p}},\boldsymbol{{is}}\:\boldsymbol{{minimum}},\boldsymbol{{when}}\:\boldsymbol{{the}}\:\boldsymbol{{points}}: \\ $$$$\boldsymbol{{B}}\:\boldsymbol{{and}}\:\:\boldsymbol{{D}},\boldsymbol{{C}},\boldsymbol{{are}}\:\boldsymbol{{colinear}}. \\ $$$$\boldsymbol{{so}},\boldsymbol{{because}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{triangle}}:\boldsymbol{{ABC}}, \\ $$$$\boldsymbol{{is}}\:\boldsymbol{{rightangle}},\boldsymbol{{then}}: \\ $$$$\boldsymbol{{BC}}=\boldsymbol{{p}}_{\boldsymbol{{min}}} =\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} }=\mathrm{4}\sqrt{\mathrm{5}}\:\:.\blacksquare \\ $$

Commented by behi834171 last updated on 25/Jul/25