Find-maximum-value-of-f-x-16-x-2-x-2-9-

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

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

Answered by EDWIN88 last updated on 02/Mar/21

let g(x)=(16−x^2 )(x^2 −9) ⇒((dg(x))/dx)=−2x(x^2 −9)+2x(16−x^2 )=0 −2x{x^2 −9−(16−x^2 )}=0 −2x(2x^2 −25)=0 → { ((x=0)),((x=± (5/( (√2))))) :} ((d^2 g(x))/dx^2 ) = −12x^2 +60 < 0 for x = ± (5/( (√2))) it follow that g(x) maximum at x= ± (5/( (√2))) so f(x)_(max) = (√((16−((25)/2))(((25)/2)−9))) = (√((7/2)×(7/2))) = (7/2) .

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

Commented by bramlexs22 last updated on 02/Mar/21

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

((a+b)/2)≥(√(ab)) f(x) = (√((16−x^2 )(x^2 −9))) ≤(1/2)(16−x^2 +x^2 −9)=(7/2) i.e. maximum=(7/2)

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

Commented by bramlexs22 last updated on 02/Mar/21

$$\mathrm{AM}−\mathrm{GM} \\ $$

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