Question

$I:$ The sides of a rectangle are $(6-x)cm$ and $(x-3)cm$. If its area is maximum, then $x=4.5$
$II:$ The maximum area of the rectangle inscribed in a circle of radius $5cm$ is $50sq.cm$. Which of the above statements are true?

• A. only $I$
• B. only $II$
• C. both $I$ and $II$
• D. neither $I$ nor $II$

Answer 1

The correct option is C both $I$ and $II$

(I) $f\left(x\right)=(6-x)(x-3)$

$f^{\prime }\left(x\right)=6-x-x+3$

$0=9-2x$

$x=4\cdot 5$

(II) $\frac{a^2}{4}+\frac{b^2}{4}=r^2$

Area of rectangle $=ab$

$\therefore$ Area $=2a\sqrt{r^2-\frac{a^2}{4}}$

$\frac{d\left(Area\right)}{d\left(a\right)}=2(\sqrt{r^2-\frac{a^2}{4}}-\frac{\frac{2a^2}{4}}{2\sqrt{r^2-\frac{a^2}{4}}})$

$0=(r^2-\frac{a^2}{4}-\frac{a^2}{4})$

$a=\sqrt{2}r$

$b=\sqrt{2}r$

Hence, Area $=2r^2=50c{m}^2$