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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-dimes

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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?
$$\mathrm{A}\:\mathrm{farmer}\:\mathrm{has}\:\mathrm{100}\:\mathrm{meters}\:\mathrm{of}\: \\ $$$$\mathrm{fencing}\:\mathrm{and}\:\mathrm{wants}\:\mathrm{to}\:\mathrm{enclose}\:\mathrm{an} \\ $$$$\mathrm{rectagular}\:\mathrm{field}\:\mathrm{along}\:\mathrm{a}\:\mathrm{river}.\:\mathrm{Thei} \\ $$$$\mathrm{rver}\:\mathrm{forms}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{rectangle}\:\mathrm{so}\:\mathrm{fencing}\:\mathrm{is}\:\mathrm{needed}\:\mathrm{onlyo} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{other}\:\mathrm{three}\:\mathrm{sides}.\:\mathrm{What}\: \\ $$$$\mathrm{dimesions}\:\mathrm{should}\:\mathrm{the}\:\mathrm{farmer}\: \\ $$$$\mathrm{chooseto}\:\mathrm{maximize}\:\mathrm{the}\:\mathrm{enclosed} \\ $$$$\mathrm{area}? \\ $$
Answered by mr W last updated on 06/Mar/25
the question is equivalent to the question how to form a rectangle with 200 meter fence such that its area is maximum. in this case the rectangle should be a square with side length 50 meter. therefore with 100 meter fence and along the river the largest rectangle area you can form is half of the square, i.e. 25×50=1250 m^2 .
$${the}\:{question}\:{is}\:{equivalent}\:{to}\:{the} \\ $$$${question}\:{how}\:{to}\:{form}\:{a}\:{rectangle}\: \\ $$$${with}\:\mathrm{200}\:{meter}\:{fence}\:{such}\:{that}\:{its}\: \\ $$$${area}\:{is}\:{maximum}.\:{in}\:{this}\:{case}\:{the} \\ $$$${rectangle}\:{should}\:{be}\:{a}\:{square}\:{with} \\ $$$${side}\:{length}\:\mathrm{50}\:{meter}.\:{therefore}\: \\ $$$${with}\:\mathrm{100}\:{meter}\:{fence}\:{and}\:{along}\:{the} \\ $$$${river}\:{the}\:{largest}\:{rectangle}\:{area}\:{you} \\ $$$${can}\:{form}\:{is}\:{half}\:{of}\:{the}\:{square},\:{i}.{e}. \\ $$$$\mathrm{25}×\mathrm{50}=\mathrm{1250}\:{m}^{\mathrm{2}} . \\ $$
Commented by mr W last updated on 06/Mar/25
if the farmer is clever, he can enclose even more area with 100 meter fence. the maximum area he can enclose along the river is a semi−circle with a area of ((100^2 )/(2π))≈1592 m^2 > 1250 m^2
$${if}\:{the}\:{farmer}\:{is}\:{clever},\:{he}\:{can} \\ $$$${enclose}\:{even}\:{more}\:{area}\:{with} \\ $$$$\mathrm{100}\:{meter}\:{fence}.\:{the}\:{maximum} \\ $$$${area}\:{he}\:{can}\:{enclose}\:{along}\:{the}\:{river} \\ $$$${is}\:{a}\:{semi}−{circle}\:{with}\:{a}\:{area}\:{of} \\ $$$$\frac{\mathrm{100}^{\mathrm{2}} }{\mathrm{2}\pi}\approx\mathrm{1592}\:{m}^{\mathrm{2}} \:>\:\mathrm{1250}\:{m}^{\mathrm{2}} \\ $$
Commented by mr W last updated on 06/Mar/25
Commented by Rasheed.Sindhi last updated on 06/Mar/25
Ni⊂∈! Thanks sir!
$$\mathbb{N}\boldsymbol{\mathrm{i}}\subset\in! \\ $$$$\mathbb{T}\mathrm{han}\Bbbk\mathrm{s}\:\mathrm{sir}! \\ $$
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)))
$$\bullet\mathrm{Let}\:{x}\:\mathrm{be}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{sideparallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{river}. \\ $$$$\bullet\mathrm{Let}\:{y}\:\mathrm{be}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{each}\: \\ $$$$\mathrm{ofthe}\:\mathrm{two}\:\mathrm{sidesr}\:\mathrm{perpendicula}\:\mathrm{to}\:\mathrm{the}\:\mathrm{river} \\ $$$$\mathrm{Total}\:\mathrm{fencing}=\mathrm{100}\left({m}\right) \\ $$$$\mathrm{2}{y}+{x}=\mathrm{100}\left(\mathrm{since}\:\mathrm{only}\:\mathrm{three}\:\mathrm{sides}\:\mathrm{needn}\right. \\ $$$$\left.\mathrm{fecing}\right) \\ $$$$\mathrm{Arwa}\:\mathrm{A}={x}+{y} \\ $$$${x}=\mathrm{100}−\mathrm{2}{y}\left(\&\mathrm{Substitute}\:\mathrm{into}\:\mathrm{area}\:\mathrm{equation}\right)\Rightarrow\mathrm{A}={y}\left(\mathrm{100}−\mathrm{2}{y}\right),{A}=\mathrm{100}{y}−\mathrm{2}{y}^{\mathrm{2}} \\ $$$$\mathrm{A} \\ $$$$\mathrm{To}\:\mathrm{find}\:\mathrm{maximum}\:\mathrm{area}\:\mathrm{takee} \\ $$$$\mathrm{drivative}\:\mathrm{and}\:\mathrm{set}\:\mathrm{to}\:\mathrm{0}: \\ $$$$\frac{{dA}}{{dy}}=\mathrm{100}−\mathrm{4}{y}=\mathrm{0} \\ $$$$\mathrm{4}{y}=\mathrm{100} \\ $$$${y}=\mathrm{25} \\ $$$$\mathrm{Substitute}\:\mathrm{back}\:\mathrm{to}\:\mathrm{find}\:{x}: \\ $$$$\begin{cases}{{x}=\mathrm{50}}\\{{y}=\mathrm{25}}\end{cases} \\ $$$$\mathrm{so}\Rightarrow\begin{array}{|c|}{\mathrm{50}{m}×\mathrm{25}{m}}\\\hline\end{array} \\ $$
Commented by Rasheed.Sindhi last updated on 06/Mar/25
Thanks sir!
$$\mathcal{T}{hanks}\:{sir}! \\ $$

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