Question

When $x$ lies between$-6$ and $8$. The maximum value of $(x+6{)}^4(8-x{)}^3$ is

• A. $6^4{8}^3$
• B. $8^4{7}^3$
• C. $8^4{6}^3$
• D. $7^4{5}^3$

Answer 1

The correct option is C $8^4{6}^3$

$f\left(x\right)=(x+6{)}^4(8-x{)}^3$
$\Rightarrow f^{\prime }\left(x\right)=4(x+6{)}^3(8-x{)}^3-3(x+6{)}^4(8-x{)}^2=7(x+6{)}^3(8-x{)}^2(2-x)=0$
$\Rightarrow x=-6,2,8$
$\Rightarrow f^{\prime \prime }\left(x\right)=7[-(x+6{)}^3(8-x{)}^2+3(x+6{)}^2(8-x{)}^2(2-x)-2(x+6{)}^3(8-x)(2-x)]$
Since, $f^{\prime \prime }(-6)=0$, $f^{\prime \prime }\left(8\right)=0$ & $f^{\prime \prime }\left(2\right)=7[-8^3{6}^2]<0$
Therefore, $f\left(x\right)$ is maximum at $x=2$
Thus, $f\left(2\right)=(2+6{)}^4(8-2{)}^3=8^4{6}^3$
Ans: C