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Question-30627




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   (dy/dx)=3(x+2)^2 (2−x)^4 −4(x+2)^3 (2−x)^3   (dy/dx)=0 for −2 < x < 2   ⇒              3(2−x)=4(x+2)     or      x = −(2/7)  y_(max)   for  x < 2 is             = (−(2/7)+2)^3 (2+(2/7))^4             = (((12)^3 (16)^4 )/((7)^7 )) = 2×((6^3 .8^6 )/7^7 ) .
$${y}=\left({x}+\mathrm{2}\right)^{\mathrm{3}} \left(\mathrm{2}−{x}\right)^{\mathrm{4}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{3}\left({x}+\mathrm{2}\right)^{\mathrm{2}} \left(\mathrm{2}−{x}\right)^{\mathrm{4}} −\mathrm{4}\left({x}+\mathrm{2}\right)^{\mathrm{3}} \left(\mathrm{2}−{x}\right)^{\mathrm{3}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{0}\:{for}\:−\mathrm{2}\:<\:{x}\:<\:\mathrm{2}\:\:\:\Rightarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\left(\mathrm{2}−{x}\right)=\mathrm{4}\left({x}+\mathrm{2}\right) \\ $$$$\:\:\:{or}\:\:\:\:\:\:{x}\:=\:−\frac{\mathrm{2}}{\mathrm{7}} \\ $$$${y}_{{max}} \:\:{for}\:\:{x}\:<\:\mathrm{2}\:{is} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\left(−\frac{\mathrm{2}}{\mathrm{7}}+\mathrm{2}\right)^{\mathrm{3}} \left(\mathrm{2}+\frac{\mathrm{2}}{\mathrm{7}}\right)^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\frac{\left(\mathrm{12}\right)^{\mathrm{3}} \left(\mathrm{16}\right)^{\mathrm{4}} }{\left(\mathrm{7}\right)^{\mathrm{7}} }\:=\:\mathrm{2}×\frac{\mathrm{6}^{\mathrm{3}} .\mathrm{8}^{\mathrm{6}} }{\mathrm{7}^{\mathrm{7}} }\:. \\ $$
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
To Ajfour sir :  we cannot  apply  Am−Gm method as here x can be −ve also.
$$\mathrm{To}\:\mathrm{Ajfour}\:\mathrm{sir}\::\:\:\mathrm{we}\:\mathrm{cannot}\:\:\mathrm{apply} \\ $$$$\mathrm{Am}−\mathrm{Gm}\:\mathrm{method}\:\mathrm{as}\:\mathrm{here}\:{x}\:{can}\:{be}\:−{ve}\:{also}. \\ $$
Commented by ajfour last updated on 23/Feb/18
good! i did not apply either.
$${good}!\:{i}\:{did}\:{not}\:{apply}\:{either}. \\ $$
Commented by Tinkutara last updated on 24/Feb/18
Even when x is −ve 2−x and 2+x  are +ve so we can apply AM-GM.
$${Even}\:{when}\:{x}\:{is}\:−{ve}\:\mathrm{2}−{x}\:{and}\:\mathrm{2}+{x} \\ $$$${are}\:+{ve}\:{so}\:{we}\:{can}\:{apply}\:{AM}-{GM}. \\ $$

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