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Question Number 30627 by Tinkutara last updated on 23/Feb/18

Answered by ajfour last updated on 23/Feb/18

$$y={(x+2)}^3{(2-x)}^4$$$$\frac{dy}{dx}=3{(x+2)}^2{(2-x)}^4-4{(x+2)}^3{(2-x)}^3$$$$\frac{dy}{dx}=0for-2<x<2\Rightarrow$$$$3(2-x)=4(x+2)$$$$orx=-\frac{2}{7}$$$$y_{max}forx<2is$$$$={(-\frac{2}{7}+2)}^3{(2+\frac{2}{7})}^4$$$$=\frac{{\left(12\right)}^3{\left(16\right)}^4}{{\left(7\right)}^7}=2\times \frac{6^3.8^6}{7^7}.$$

Commented by Tinkutara last updated on 23/Feb/18

Ohh sorry that was the same answer

Commented by rahul 19 last updated on 23/Feb/18

$$\mathrm{To}\mathrm{Ajfour}\mathrm{sir}:\mathrm{we}\mathrm{cannot}\mathrm{apply}$$$$\mathrm{Am}-\mathrm{Gm}\mathrm{method}\mathrm{as}\mathrm{here}xcanbe-vealso.$$

Commented by ajfour last updated on 23/Feb/18

$$good!ididnotapplyeither.$$

Commented by Tinkutara last updated on 24/Feb/18

$$Evenwhenxis-ve2-xand2+x$$$$are+vesowecanapplyAM-GM.$$

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