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

$${Find}\:{the}\:{dimensions}\:{of}\:{the} \\ $$$${largest}\:{rectangular}\:{garden}\:{that} \\ $$$${can}\:{be}\:{enclosed}\:{by}\:\mathrm{80}{m}\:{of} \\ $$$${fencing} \\ $$

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

$$\mathrm{2}\left({l}+{b}\right)=\mathrm{80} \\ $$$${l}+{b}=\mathrm{40} \\ $$$${A}={lb} \\ $$$${A}={l}\left(\mathrm{40}−{l}\right)\rightarrow{A}=\mathrm{40}{l}−{l}^{\mathrm{2}} \\ $$$$\frac{{dA}}{{dl}}=\mathrm{40}−\mathrm{2}{l} \\ $$$${for}\:{max}/{min}\:\frac{{dA}}{{dl}}=\mathrm{0}\rightarrow \\ $$$$\mathrm{40}−\mathrm{2}{l}=\mathrm{0}\rightarrow{l}=\mathrm{20} \\ $$$${b}=\mathrm{40}−\mathrm{20}=\mathrm{20} \\ $$$$ \\ $$$$\frac{{d}^{\mathrm{2}} {A}}{{dl}^{\mathrm{2}} }=−\mathrm{2}\:<\mathrm{0} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{length}}=\mathrm{20} \\ $$$$\boldsymbol{{breath}}=\mathrm{20} \\ $$$$ \\ $$

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

$${thank}\:{you}\:{very}\:{much} \\ $$

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

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

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