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Question Number 151685 by iloveisrael last updated on 22/Aug/21

$$Findmaximumvalueoffunction$$$$\alpha \left(x\right)=\sqrt{2x}+\sqrt{16-x}+\sqrt{35+x}.$$

Answered by mr W last updated on 22/Aug/21

$$x⩾0$$$$x⩽16$$$$f^{\prime }\left(x\right)=\frac{1}{\sqrt{2x}}-\frac{1}{2\sqrt{16-x}}+\frac{1}{2\sqrt{35+x}}=0$$$$\frac{2}{\sqrt{2x}}+\frac{1}{\sqrt{35+x}}=\frac{1}{\sqrt{16-x}}$$$$\frac{2}{x}+\frac{1}{35+x}+\frac{2\sqrt{2}}{\sqrt{x(35+x)}}=\frac{1}{16-x}$$$$\frac{2\sqrt{2}}{\sqrt{x(35+x)}}=\frac{x(35+x)-x(16-x)-2(35+x)(16-x)}{(16-x)\left(x\right)(35+x)}$$$$\frac{2\sqrt{2}}{\sqrt{x(35+x)}}=\frac{4x^2+57x-1120}{(16-x)\left(x\right)(35+x)}$$$$8=\frac{{(4x^2+57x-1120)}^2}{{(16-x)}^2\left(x\right)(35+x)}$$$$8x^4+432x^3+1201x^2-199360x+1254400=0$$$$\Rightarrow x\approx 12.6106$$$$f{\left(x\right)}_{max}\approx 13.7631$$

Commented by mr W last updated on 22/Aug/21

$$f{\left(x\right)}_{min}=f\left(0\right)=4+\sqrt{35}$$

Commented by MJS_new last updated on 22/Aug/21

$$\mathrm{I}\mathrm{also}\mathrm{tried}.\mathrm{no}\mathrm{useable}\mathrm{exact}\mathrm{solution}\mathrm{possible}$$

Commented by mr W last updated on 22/Aug/21

$$soitis,sir.ifoundnoexactway.$$

Commented by iloveisrael last updated on 22/Aug/21

$$forminimumvalue,igotexactvalue$$

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