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

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=lb A=l(40−l)→A=40l−l^2 (dA/dl)=40−2l for max/min (dA/dl)=0→ 40−2l=0→l=20 b=40−20=20 (d^2 A/dl^2 )=−2 <0 so length=20 breath=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⇒a+b=40 and the area of the garden equal to S=ab=a(40−a) =−a^2 +40a=−(a−20)^2 +400≤400 ⇒S≤400 which means S_(max) =400 m^2 when a=b=20m

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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