A-rectangular-enclosure-is-to-be-made-against-a-straight-wall-using-three-lengths-of-fencing-The-total-length-of-the-fencing-available-is-50m-Show-that-the-area-of-the-enclosure-is-50x-2x-2-wher

Question Number 204826 by necx122 last updated on 28/Feb/24

A r e c t a n g u l a r e n c l o s u r e i s t o b e m a d e

a g a i n s t a s t r a i g h t w a l l u s i n g t h r e e

l e n g t h s o f f e n c i n g . T h e t o t a l l e n g t h o f

t h e f e n c i n g a v a i l a b l e i s 50 m . S h o w

t h a t t h e a r e a o f t h e e n c l o s u r e i s

50 x - 2 x^{2}, w h e r e x i s t h e l e n g t h o f t h e

s i d e s p e r p e n d i c u l a r t o t h e w a l l . H e n c e

f i n d t h e m a x i m u m a r e a o f t h e

e n c l o s u r e .

Commented by mr W last updated on 29/Feb/24

w e g e t

R = \sqrt{\frac{a^{2} + b^{2} + c^{2}}{3}} \sin {\frac{π}{3} + \frac{1}{3} \sin^{- 1} [a b c {(\frac{3}{a^{2} + b^{2} + c^{2}})}^{\frac{3}{2}}]}

= \sqrt{\frac{10^{2} + 15^{2} + 25^{2}}{3}} \sin {\frac{π}{3} + \frac{1}{3} \sin^{- 1} [10 \times 15 \times 25 \times {(\frac{3}{10^{2} + 15^{2} + 25^{2}})}^{\frac{3}{2}}]}

\approx 17.0977

A D = d = 2 R \approx 34.1594

s = \frac{10 + 15 + 25 + 34.1594}{2} = 42.0977

{[A B C D]}_{m a x} = \sqrt{(42.0977 - 10) (42.0977 - 15) (42.0977 - 25) (42.0977 - 34.1594)}

= 343.586 m^{2}

Commented by mr W last updated on 29/Feb/24

i t i s f o u n d o u t t h a t t h e a r e a o f t h e

e n c l o s u r e i s m a x i m u m w h e n t h e

f e n c e f o l l o w s a c i r c l e w i t h d i a m e t e r

o n t h e w a l l .

Commented by mr W last updated on 29/Feb/24

Commented by mr W last updated on 29/Feb/24

t h e a b s o l u t e m a x i m u m a r e a w e c a n

g e t w i t h a t o t a l l e n g t h o f f e n c e o f

50 m i s w h e n t h e f e n c e f o r m s a

s e m i - c i r c l e .

R = \frac{50}{π}

{A r e a}_{m a x} = \frac{π R^{2}}{2} = \frac{50^{2}}{2 π} \approx 397.887 m^{2}

Commented by mr W last updated on 28/Feb/24

a m o r e c h a l l e n g i n g v e r s i o n o f t h e

q u e s t i o n :

i f t h e t h r e e l e n g t h e s o f t h e f e n c i n g

a r e 10 m, 15 m a n d 25 m r e s p e c t i v e l y .

w h a t i s t h e l a r g e s t a r e a o f t h e

e n c l o s u r e w e c a n g e t ?

Commented by necx122 last updated on 28/Feb/24

{W e}^{'} l l a w a i t y o u r b e a u t i f u l s o l u t i o n t o

t h i s a s y o u s o l v e d e a r M r . W

Commented by mr W last updated on 29/Feb/24

w e c a n s e e t h i s i s l a r g e r t h a n t h e

m a x i m u m a r e a o f t h e r e c t a n g u l a r

e n c l o s u r e 312.5 m^{2} .

Commented by necx122 last updated on 29/Feb/24

My goodness! I've never encountered that formula you used. Thank you, I'll research it.

Commented by mr W last updated on 29/Feb/24

i h a v e d e r i v e d t h a t f o r m u l a b y m y s e l f .

s o i a m n o t s u r e i f y o u c a n f i n d i t

s o m e w h e r e e l s e . p l e a s e l e t m e k n o w

w h a t y o u f o u n d i n y o u r r e s e a r c h .

Commented by MathematicalUser2357 last updated on 04/Mar/24

\sqrt{\frac{10^{2} + 15^{2} - 25^{2}}{3}} \sin [\frac{π}{3} + \frac{1}{3} \sin^{- 1} {10 \times 15 \times 25 \times {(\frac{3}{10^{2} + 15^{2} + 25^{2}})}^{\frac{3}{2}}}] ? ? ?

Commented by mr W last updated on 04/Mar/24

w h a t d o y o u w a n t t o a s k c o n c r e t e l y ?

Answered by Rasheed.Sindhi last updated on 28/Feb/24

L e t x, y a r e t w o d i m e n t i o n s

T h r e e s i d e s =

2 x + y = 50 \Rightarrow y = 50 - 2 x

A r e a o f e n c l o s u r e

= x y = x (50 - 2 x) = 50 x - 2 x^{2}

Commented by necx122 last updated on 28/Feb/24

T h a n k y o u s i r . I c a n r e a d i l y p r o c e e d

f r o m h e r e .

Commented by mr W last updated on 29/Feb/24

m a x i m u m = 2 \times 12 . 5^{2} = 312.5 m^{2}

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