+483 Let x, y, and z be real numbers. If x^2 + y^2 + z^2 = 1, then find the maximum value of
3x + 4y + 5z + x^3 + \frac{4x^2 y}{z} + \frac{z^5}{xy^2}
+15127 \(0\)
\(Find\ the\ maximum\ value\ of\ \ 3x + 4y + 5z + x^3 + \dfrac{4x^2 y}{z} + \dfrac{z^5}{xy^2}.\)
Plane geometry
\(z^2=x^2+y^2\ \color{green }Pythagorean\ theorem\\ \color{blue}y=x\\ z^2=2x^2\\ \color{blue}z=x\sqrt{2}\\ x^2+y^2+z^2=1\\ x^2+x^2+2x^2=1\\\)
\(4x^2=1\\ \color{blue}x=\dfrac{1}{2}\)
\(f(x)=3x + 4y + 5z + x^3 + \dfrac{4x^2 y}{z} + \dfrac{z^5}{xy^2}\\ f(x)=3x+4x+5x\sqrt{2}+x^3+\dfrac{4x^3}{x\sqrt{2}}+\dfrac{(x\sqrt{2})^5}{x^3}\\ f(x)=x^3+6\sqrt{2}x^2+5\sqrt{2}x+7x\ \color{green}according\ to\ WolframAlpha\\ f(x_{0.5})=0.5^3+6\sqrt{2}(0.5)^2+5\sqrt{2}(0.5)+7(0.5)=9.28185\\ \dfrac{d}{dx}=3x^2+3⋅2^{\frac{5}{2}}x+5√2+7=0\\ {\color{blue}x\in \{-4.6477};-1.009\}\)
\(f(x)_{max}=(-4.6477)^3+6\sqrt{2}(-4.6477)^2+5\sqrt{2}(-4.6477)+7(-4.6477)\\ \color{blue}f(x)_{max}=17.4979\)
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