A farmer has 100 meters of fencing and wants to enclose an rectagular field along a river. Thei rver forms one side of the rectangle so fencing is needed onlyo for the other three sides. What dimesions should the farmer chooseto maximize the enclosed area?


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Question Number 217203 by Rasheed.Sindhi last updated on 05/Mar/25

$A farmer has 100 meters of$ $fencing and wants to enclose an$ $rectagular field along a river . Thei$ $rver forms one side of the$ $rectangle so fencing is needed onlyo$ $for the other three sides . What$ $dimesions should the farmer$ $chooseto maximize the enclosed$ $area ?$
	Answered by mr W last updated on 06/Mar/25

	$t h e q u e s t i o n i s e q u i v a l e n t t o t h e$ $q u e s t i o n h o w t o f o r m a r e c t a n g l e$ $w i t h 200 m e t e r f e n c e s u c h t h a t i t s$ $a r e a i s m a x i m u m . i n t h i s c a s e t h e$ $r e c t a n g l e s h o u l d b e a s q u a r e w i t h$ $s i d e l e n g t h 50 m e t e r . t h e r e f o r e$ $w i t h 100 m e t e r f e n c e a n d a l o n g t h e$ $r i v e r t h e l a r g e s t r e c t a n g l e a r e a y o u$ $c a n f o r m i s h a l f o f t h e s q u a r e, i . e .$ $25 \times 50 = 1250 m^{2} .$
	Commented by mr W last updated on 06/Mar/25

	$i f t h e f a r m e r i s c l e v e r, h e c a n$ $e n c l o s e e v e n m o r e a r e a w i t h$ $100 m e t e r f e n c e . t h e m a x i m u m$ $a r e a h e c a n e n c l o s e a l o n g t h e r i v e r$ $i s a s e m i - c i r c l e w i t h a a r e a o f$ $\frac{100^{2}}{2 π} \approx 1592 m^{2} > 1250 m^{2}$
	Commented by mr W last updated on 06/Mar/25

	Commented by Rasheed.Sindhi last updated on 06/Mar/25

	$N i \subset\in!$ $T han k s sir!$
	Answered by MrGaster last updated on 06/Mar/25
	$•Let x be the length of the sideparallel to the river. •Let y be the length of each ofthe two sidesr perpendicula to the river Total fencing=100(m) 2y+x=100(since only three sides needn fecing) Arwa A=x+y x=100−2y(&Substitute into area equation)⇒A=y(100−2y),A=100y−2y^2 A To find maximum area takee drivative and set to 0: (dA/dy)=100−4y=0 4y=100 y=25 Substitute back to find x: { ((x=50)),((y=25)) :} so⇒ determinant (((50m×25m)))$
	$∙ Let x be the length of the$ $sideparallel to the river .$ $∙ Let y be the length of each$ $ofthe two sidesr perpendicula to the river$ $Total fencing = 100 (m)$ $2 y + x = 100 (since only three sides needn$ $fecing)$ $Arwa A = x + y$ $x = 100 - 2 y (& Substitute into area equation) \Rightarrow A = y (100 - 2 y), A = 100 y - 2 y^{2}$ $A$ $To find maximum area takee$ $drivative and set to 0 :$ $\frac{d A}{d y} = 100 - 4 y = 0$ $4 y = 100$ $y = 25$ $Substitute back to find x :$ ${\begin{cases} x = 50 \\ y = 25 \end{cases}$ $so \Rightarrow \begin{matrix} 50 m \times 25 m \end{matrix}$
	Commented by Rasheed.Sindhi last updated on 06/Mar/25

	$T h a n k s s i r!$
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