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Question Number 134285 by bramlexs22 last updated on 02/Mar/21

$$\mathrm{Find}\mathrm{maximum}\mathrm{value}\mathrm{of}$$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{(16-{\mathrm{x}}^2)({\mathrm{x}}^2-9)}$$

Answered by EDWIN88 last updated on 02/Mar/21

$$\mathrm{let}\mathrm{g}\left(\mathrm{x}\right)=(16-{\mathrm{x}}^2)({\mathrm{x}}^2-9)\Rightarrow \frac{\mathrm{dg}\left(\mathrm{x}\right)}{\mathrm{dx}}=-2\mathrm{x}({\mathrm{x}}^2-9)+2\mathrm{x}(16-{\mathrm{x}}^2)=0$$$$-2\mathrm{x}\{{\mathrm{x}}^2-9-(16-{\mathrm{x}}^2)\}=0$$$$-2\mathrm{x}({2\mathrm{x}}^2-25)=0\to \{\begin{array}{l}\mathrm{x}=0\\ \mathrm{x}=\pm \frac{5}{\sqrt{2}}\end{array}$$$$\frac{{\mathrm{d}}^2\mathrm{g}\left(\mathrm{x}\right)}{{\mathrm{dx}}^2}=-{12\mathrm{x}}^2+60<0\mathrm{for}\mathrm{x}=\pm \frac{5}{\sqrt{2}}$$$$\mathrm{it}\mathrm{follow}\mathrm{that}\mathrm{g}\left(\mathrm{x}\right)\mathrm{maximum}\mathrm{at}\mathrm{x}=\pm \frac{5}{\sqrt{2}}$$$$\mathrm{so}\mathrm{f}{\left(\mathrm{x}\right)}_{\max }=\sqrt{(16-\frac{25}{2})(\frac{25}{2}-9)}$$$$=\sqrt{\frac{7}{2}\times \frac{7}{2}}=\frac{7}{2}.$$

Commented by bramlexs22 last updated on 02/Mar/21

Answered by mr W last updated on 02/Mar/21

$$\frac{a+b}{2}⩾\sqrt{ab}$$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{(16-{\mathrm{x}}^2)({\mathrm{x}}^2-9)}$$$$⩽\frac{1}{2}(16-x^2+x^2-9)=\frac{7}{2}$$$$i.e.maximum=\frac{7}{2}$$

Commented by bramlexs22 last updated on 02/Mar/21

$$\mathrm{AM}-\mathrm{GM}$$

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