Find the dimensions of the largest rectangular garden that can be enclosed by 80m of fencing


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Question Number 118570 by rexfordattacudjoe last updated on 18/Oct/20

$F i n d t h e d i m e n s i o n s o f t h e$ $l a r g e s t r e c t a n g u l a r g a r d e n t h a t$ $c a n b e e n c l o s e d b y 80 m o f$ $f e n c i n g$
	Answered by TANMAY PANACEA last updated on 18/Oct/20

	$2 (l + b) = 80$ $l + b = 40$ $A = l b$ $A = l (40 - l) \to A = 40 l - l^{2}$ $\frac{d A}{d l} = 40 - 2 l$ $f o r m a x / m i n \frac{d A}{d l} = 0 \to$ $40 - 2 l = 0 \to l = 20$ $b = 40 - 20 = 20$ $\frac{d^{2} A}{{d l}^{2}} = - 2 < 0$ $s o l e n g t h = 20$ $b r e a t h = 20$
	Commented by JDamian last updated on 18/Oct/20
	if you replace 80 with P (perimeter) you'll get P/4; which it means "square"... because the square is the rectangle which gets the biggest area for a given perimeter.
	Commented by rexfordattacudjoe last updated on 18/Oct/20

	$t h a n k y o u v e r y m u c h$
	Answered by 1549442205PVT last updated on 18/Oct/20

	$Denote by a and b be the length and$ $width of the garden respectively . Then$ $from the hypothesis we have :$ $2 (a + b) = 80 \Rightarrow a + b = 40 and the area of$ $the garden equal to S = ab = a (40 - a)$ $= - a^{2} + 40 a = - {(a - 20)}^{2} + 400 ⩽ 400$ $\Rightarrow S ⩽ 400 which means S_{\max} = 400 m^{2}$ $when a = b = 20 m$
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