Question Number 204826 by necx122 last updated on 28/Feb/24
$${A}\:{rectangular}\:{enclosure}\:{is}\:{to}\:{be}\:{made} \\ $$$${against}\:{a}\:{straight}\:{wall}\:{using}\:{three} \\ $$$${lengths}\:{of}\:{fencing}.\:{The}\:{total}\:{length}\:{of} \\ $$$${the}\:{fencing}\:{available}\:{is}\:\mathrm{50}{m}.\:{Show} \\ $$$${that}\:{the}\:{area}\:{of}\:{the}\:{enclosure}\:{is} \\ $$$$\mathrm{50}{x}\:−\:\mathrm{2}{x}^{\mathrm{2}} ,\:{where}\:{x}\:{is}\:{the}\:{length}\:{of}\:{the} \\ $$$${sides}\:{perpendicular}\:{to}\:{the}\:{wall}.\:{Hence} \\ $$$${find}\:{the}\:{maximum}\:{area}\:{of}\:{the} \\ $$$${enclosure}. \\ $$
Commented by mr W last updated on 29/Feb/24
![we get R=(√((a^2 +b^2 +c^2 )/3)) sin {(π/3)+(1/3)sin^(−1) [abc((3/(a^2 +b^2 +c^2 )))^(3/2) ]} =(√((10^2 +15^2 +25^2 )/3)) sin {(π/3)+(1/3)sin^(−1) [10×15×25×((3/(10^2 +15^2 +25^2 )))^(3/2) ]} ≈17.0977 AD=d=2R≈34.1594 s=((10+15+25+34.1594)/2)=42.0977 [ABCD]_(max) =(√((42.0977−10)(42.0977−15)(42.0977−25)(42.0977−34.1594))) =343.586 m^2](https://www.tinkutara.com/question/Q204859.png)
$${we}\:{get} \\ $$$${R}=\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{3}}}\:\mathrm{sin}\:\left\{\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \left[{abc}\left(\frac{\mathrm{3}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]\right\} \\ $$$$\:\:\:=\sqrt{\frac{\mathrm{10}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} +\mathrm{25}^{\mathrm{2}} }{\mathrm{3}}}\:\mathrm{sin}\:\left\{\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \left[\mathrm{10}×\mathrm{15}×\mathrm{25}×\left(\frac{\mathrm{3}}{\mathrm{10}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} +\mathrm{25}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]\right\} \\ $$$$\:\:\:\approx\mathrm{17}.\mathrm{0977} \\ $$$${AD}={d}=\mathrm{2}{R}\approx\mathrm{34}.\mathrm{1594} \\ $$$${s}=\frac{\mathrm{10}+\mathrm{15}+\mathrm{25}+\mathrm{34}.\mathrm{1594}}{\mathrm{2}}=\mathrm{42}.\mathrm{0977} \\ $$$$\left[{ABCD}\right]_{{max}} =\sqrt{\left(\mathrm{42}.\mathrm{0977}−\mathrm{10}\right)\left(\mathrm{42}.\mathrm{0977}−\mathrm{15}\right)\left(\mathrm{42}.\mathrm{0977}−\mathrm{25}\right)\left(\mathrm{42}.\mathrm{0977}−\mathrm{34}.\mathrm{1594}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{343}.\mathrm{586}\:{m}^{\mathrm{2}} \\ $$
Commented by mr W last updated on 29/Feb/24
$${it}\:{is}\:{found}\:{out}\:{that}\:{the}\:{area}\:{of}\:{the} \\ $$$${enclosure}\:{is}\:{maximum}\:{when}\:{the} \\ $$$${fence}\:{follows}\:{a}\:{circle}\:{with}\:{diameter} \\ $$$${on}\:{the}\:{wall}. \\ $$
Commented by mr W last updated on 29/Feb/24
Commented by mr W last updated on 29/Feb/24
$${the}\:{absolute}\:{maximum}\:{area}\:{we}\:{can} \\ $$$${get}\:{with}\:{a}\:{total}\:{length}\:{of}\:{fence}\:{of} \\ $$$$\mathrm{50}\:{m}\:{is}\:{when}\:{the}\:{fence}\:{forms}\:{a} \\ $$$${semi}−{circle}. \\ $$$${R}=\frac{\mathrm{50}}{\pi} \\ $$$${Area}_{{max}} =\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{50}^{\mathrm{2}} }{\mathrm{2}\pi}\approx\mathrm{397}.\mathrm{887}\:{m}^{\mathrm{2}} \\ $$
Commented by mr W last updated on 28/Feb/24
$${a}\:{more}\:{challenging}\:{version}\:{of}\:{the} \\ $$$${question}: \\ $$$${if}\:{the}\:{three}\:{lengthes}\:{of}\:{the}\:{fencing} \\ $$$${are}\:\mathrm{10}{m},\:\mathrm{15}{m}\:{and}\:\mathrm{25}{m}\:{respectively}. \\ $$$${what}\:{is}\:{the}\:{largest}\:{area}\:{of}\:{the} \\ $$$${enclosure}\:{we}\:{can}\:{get}? \\ $$
Commented by necx122 last updated on 28/Feb/24
$${We}'{ll}\:{await}\:{your}\:{beautiful}\:{solution}\:{to} \\ $$$${this}\:{as}\:{you}\:{solve}\:{dear}\:{Mr}.\:{W} \\ $$
Commented by mr W last updated on 29/Feb/24
$${we}\:{can}\:{see}\:{this}\:{is}\:{larger}\:{than}\:{the} \\ $$$${maximum}\:{area}\:{of}\:{the}\:{rectangular} \\ $$$${enclosure}\:\mathrm{312}.\mathrm{5}\:{m}^{\mathrm{2}} . \\ $$
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
$${i}\:{have}\:{derived}\:{that}\:{formula}\:{by}\:{myself}. \\ $$$${so}\:{i}\:{am}\:{not}\:{sure}\:{if}\:{you}\:{can}\:{find}\:{it} \\ $$$${somewhere}\:{else}.\:{please}\:{let}\:{me}\:{know} \\ $$$${what}\:{you}\:{found}\:{in}\:{your}\:{research}. \\ $$
Commented by MathematicalUser2357 last updated on 04/Mar/24
![(√((10^2 +15^2 −25^2 )/3))sin[(π/3)+(1/3)sin^(−1) {10×15×25×((3/(10^2 +15^2 +25^2 )))^(3/2) }]???](https://www.tinkutara.com/question/Q204973.png)
$$\sqrt{\frac{\mathrm{10}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} −\mathrm{25}^{\mathrm{2}} }{\mathrm{3}}}\mathrm{sin}\left[\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \left\{\mathrm{10}×\mathrm{15}×\mathrm{25}×\left(\frac{\mathrm{3}}{\mathrm{10}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} +\mathrm{25}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right\}\right]??? \\ $$
Commented by mr W last updated on 04/Mar/24
$${what}\:{do}\:{you}\:{want}\:{to}\:{ask}\:{concretely}? \\ $$
Answered by Rasheed.Sindhi last updated on 28/Feb/24
$${Let}\:{x},\:{y}\:{are}\:{two}\:{dimentions} \\ $$$${Three}\:{sides}= \\ $$$$\mathrm{2}{x}+{y}=\mathrm{50}\Rightarrow{y}=\mathrm{50}−\mathrm{2}{x} \\ $$$${Area}\:{of}\:{enclosure} \\ $$$$={xy}={x}\left(\mathrm{50}−\mathrm{2}{x}\right)=\mathrm{50}{x}−\mathrm{2}{x}^{\mathrm{2}} \\ $$
Commented by necx122 last updated on 28/Feb/24
$${Thank}\:{you}\:{sir}.\:{I}\:{can}\:{readily}\:{proceed} \\ $$$${from}\:{here}. \\ $$
Commented by mr W last updated on 29/Feb/24
$${maximum}=\mathrm{2}×\mathrm{12}.\mathrm{5}^{\mathrm{2}} =\mathrm{312}.\mathrm{5}\:{m}^{\mathrm{2}} \\ $$