Math Question and Answers


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
Question Number 155203 by nadovic last updated on 26/Sep/21

	Answered by TheHoneyCat last updated on 29/Sep/21

	$let O be the center of the circle$ $let A be a vertice of the rectangle$ $let θ be an angle (in radian) between one of$ $the two axis (O z or O y) and the (OA) line$ $The dimentions of the rectangle are$ $(2 \cos θ, 2 \sin θ) (o r t h e o p p o s i t e d e p e n d i n g$ $o n y o u r d e f i n i t i o n)$ $Hence the area A of the rectangle is :$ $A = 4 ∣ \cos θ . \sin θ ∣$ $let a = \cos \times \sin$ $∣ a ∣ is a π / 2 periodical function$ $also on [0, π / 2] ∣ a ∣= a so let us sudy the$ $restriction of a to this set :$ $\frac{d a}{d θ} = \frac{d \cos}{d θ} \sin + \frac{d \sin}{d θ} \cos$ $= - \sin^{2} + \cos^{2}$ $so on this restriction$ $a^{'} (θ) = 0 \Leftrightarrow {(\cos θ)}^{2} = {(\sin θ)}^{2}$ $\Leftrightarrow \cos θ = \sin θ ({c a u s}^{'} t h e y a r e p o s i t i v o n t h e$ $i n t e r v a l)$ $\Leftrightarrow θ =^{π} /_{4}$ $\Leftrightarrow \cos θ = \sin θ = \sqrt{2} / 2$ $\Leftrightarrow a (θ) = \frac{1}{2}$ $Hence, the maximum area of the rectangle$ $is that of a square i e :$ $A = 2$
	Commented by nadovic last updated on 29/Sep/21

	$S i r, t h e a r e a i s g i v e n a s 32 s q . i n .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum