Question

The maximum value of ${(\sqrt{-3+4x-x^2}+4)}^2+{(x-5)}^2$, (where $1\le x\le 3$) is

• A. $34$
• B. $36$
• C. $32$
• D. $20$

Answer 1

The correct option is B $36$

$f\left(x\right)=-3+4x-x^2+16+8\sqrt{-3+4x-x^2}+x^2+25-10x$

$f\left(x\right)=38-6x+8\sqrt{-3+4x-x^2}$

$f^{\prime }\left(x\right)=0$

$-6+\frac{8(4-2x)}{2\sqrt{-3+4x-x^2}}=0$

$\frac{2(4-2x)}{\sqrt{-3+4x-x^2}}=3$

$4(4-2x{)}^2=9(-3+4x-x^2)$

$64+16x^2-64x=-27+36x-9x^2$

$25x^2-100x+91=0$

$25x^2-65x-35x+91=0$

$5x(5x-13)-7(5x-13)=0$

$x=\frac{7}{5},x=\frac{13}{5}$

mass $x=\frac{7}{5}$

Mass value$=$

${(\sqrt{-3+4.\frac{7}{5}-{\left(\frac{7}{5}\right)}^2}+4)}^2+{\left(\frac{7}{5}.5\right)}^2$

${(\sqrt{-3+\frac{28}{5}-\frac{49}{25}}+4)}^2+{\left(\frac{18}{5}\right)}^2$

${(\sqrt{\frac{10-49-75}{25}}+4)}^2+{\left(\frac{18}{5}\right)}^2$

${(\frac{4}{5}+4)}^2+{\left(\frac{18}{5}\right)}^2$

${\left(\frac{24}{5}\right)}^2+{\left(\frac{18}{5}\right)}^2$

$=\frac{576+324}{25}$

$=\frac{900}{25}$

$=36$.